Can you help me solve this algebra problem?8th Grade Algebra Problem I Can't SolveA problem with exponent...

Crack the bank account's password!

Is the tritone (A4 / d5) still banned in Roman Catholic music?

Why write a book when there's a movie in my head?

Is there any danger of my neighbor having my wife's signature?

In a post apocalypse world, with no power and few survivors, would Satnav still work?

In the Lost in Space intro why was Dr. Smith actor listed as a special guest star?

Coworker asking me to not bring cakes due to self control issue. What should I do?

Is it possible to detect 100% of SQLi with a simple regex?

Why can all solutions to the simple harmonic motion equation be written in terms of sines and cosines?

How to regain lost focus?

Neglect higher order derivatives in expression

Could a civilization in medieval fantasy world sustain itself by occupying important trade hubs and taxing trade?

What is wrong with my use of "find -print0"?

How can guns be countered by melee combat without raw-ability or exceptional explanations?

What could cause an entire planet of humans to become aphasic?

Is there a way to pause a running process on Linux systems and resume later?

Linearity Assumption

Words of Worship and Nefarious Lich

What really causes series inductance of capacitors?

Will the duration of traveling to Ceres using the same tech developed for going to Mars be proportional to the distance to go to Mars or not?

How bad is a Computer Science course that doesn't teach Design Patterns?

What does an unprocessed RAW file look like?

Do the speed limit reductions due to pollution also apply to electric cars in France?

Isn't a semicolon (';') needed after a function declaration in C++?



Can you help me solve this algebra problem?


8th Grade Algebra Problem I Can't SolveA problem with exponent laws…How can I solve for this variable?Age Problem (Equation Writing Help)Help on Algebra ProblemCan someone show me how to solve these matrix?Precalculus algebra problemIf $y=frac {a+bz}{c+dz}$, $z=frac{a+bx}{c+dx}$, $x=frac{a+by}{c+dy}$, then $ad + bc + b^2 + c^2 = 0$Algebra precalculus problemPrecalculus algebra exercise













6












$begingroup$


Hi I need to solve this problem and I don’t know how. I’d appreciate your help.




If $x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}$ and $xneq yneq z$, then $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$




I think I need to



$x^3 - ayz = x^2k$



$y^3 - azx = y^2k,$



$z^3 - axy = z^2k$



then to multiply both sides of each equality by some quantity, add them all together and factor but I don’t know how to find that quantity.






algebra-precalculus systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you trying to prove the last equality? It's not clear.
    $endgroup$
    – Victor S.
    3 hours ago










  • $begingroup$
    @VictorS. I need to prove that each expression is equal to x + y + z - a
    $endgroup$
    – questions about math
    3 hours ago












  • $begingroup$
    This isn't an answer. However, it could help... $$$$Observe that $$(x+y+z-a)-bigg(x-frac{ayz}{x^2}bigg)=frac{ayz+yx^2+zx^2-ax^2}{x^2}=frac{a(yz-x^2)+x^2(y+z)}{x^2}$$ This equals zero iff $$a(x^2-yz)=x^2(y+z)iff a-frac{ayz}{x^2}=y+z$$ $$iff a·bigg(frac{x^2-yz}{x^2}bigg)=y+ziff a=frac{x^2y+x^2z}{x^2-yz}$$
    $endgroup$
    – Dr. Mathva
    1 hour ago


















6












$begingroup$


Hi I need to solve this problem and I don’t know how. I’d appreciate your help.




If $x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}$ and $xneq yneq z$, then $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$




I think I need to



$x^3 - ayz = x^2k$



$y^3 - azx = y^2k,$



$z^3 - axy = z^2k$



then to multiply both sides of each equality by some quantity, add them all together and factor but I don’t know how to find that quantity.






algebra-precalculus systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you trying to prove the last equality? It's not clear.
    $endgroup$
    – Victor S.
    3 hours ago










  • $begingroup$
    @VictorS. I need to prove that each expression is equal to x + y + z - a
    $endgroup$
    – questions about math
    3 hours ago












  • $begingroup$
    This isn't an answer. However, it could help... $$$$Observe that $$(x+y+z-a)-bigg(x-frac{ayz}{x^2}bigg)=frac{ayz+yx^2+zx^2-ax^2}{x^2}=frac{a(yz-x^2)+x^2(y+z)}{x^2}$$ This equals zero iff $$a(x^2-yz)=x^2(y+z)iff a-frac{ayz}{x^2}=y+z$$ $$iff a·bigg(frac{x^2-yz}{x^2}bigg)=y+ziff a=frac{x^2y+x^2z}{x^2-yz}$$
    $endgroup$
    – Dr. Mathva
    1 hour ago
















6












6








6


1



$begingroup$


Hi I need to solve this problem and I don’t know how. I’d appreciate your help.




If $x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}$ and $xneq yneq z$, then $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$




I think I need to



$x^3 - ayz = x^2k$



$y^3 - azx = y^2k,$



$z^3 - axy = z^2k$



then to multiply both sides of each equality by some quantity, add them all together and factor but I don’t know how to find that quantity.






algebra-precalculus systems-of-equations







share|cite|improve this question











$endgroup$




Hi I need to solve this problem and I don’t know how. I’d appreciate your help.




If $x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}$ and $xneq yneq z$, then $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$




I think I need to



$x^3 - ayz = x^2k$



$y^3 - azx = y^2k,$



$z^3 - axy = z^2k$



then to multiply both sides of each equality by some quantity, add them all together and factor but I don’t know how to find that quantity.






algebra-precalculus systems-of-equations




algebra-precalculus systems-of-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 35 mins ago









Michael Rozenberg

104k1892197




104k1892197










asked 3 hours ago









questions about mathquestions about math

644




644












  • $begingroup$
    Are you trying to prove the last equality? It's not clear.
    $endgroup$
    – Victor S.
    3 hours ago










  • $begingroup$
    @VictorS. I need to prove that each expression is equal to x + y + z - a
    $endgroup$
    – questions about math
    3 hours ago












  • $begingroup$
    This isn't an answer. However, it could help... $$$$Observe that $$(x+y+z-a)-bigg(x-frac{ayz}{x^2}bigg)=frac{ayz+yx^2+zx^2-ax^2}{x^2}=frac{a(yz-x^2)+x^2(y+z)}{x^2}$$ This equals zero iff $$a(x^2-yz)=x^2(y+z)iff a-frac{ayz}{x^2}=y+z$$ $$iff a·bigg(frac{x^2-yz}{x^2}bigg)=y+ziff a=frac{x^2y+x^2z}{x^2-yz}$$
    $endgroup$
    – Dr. Mathva
    1 hour ago




















  • $begingroup$
    Are you trying to prove the last equality? It's not clear.
    $endgroup$
    – Victor S.
    3 hours ago










  • $begingroup$
    @VictorS. I need to prove that each expression is equal to x + y + z - a
    $endgroup$
    – questions about math
    3 hours ago












  • $begingroup$
    This isn't an answer. However, it could help... $$$$Observe that $$(x+y+z-a)-bigg(x-frac{ayz}{x^2}bigg)=frac{ayz+yx^2+zx^2-ax^2}{x^2}=frac{a(yz-x^2)+x^2(y+z)}{x^2}$$ This equals zero iff $$a(x^2-yz)=x^2(y+z)iff a-frac{ayz}{x^2}=y+z$$ $$iff a·bigg(frac{x^2-yz}{x^2}bigg)=y+ziff a=frac{x^2y+x^2z}{x^2-yz}$$
    $endgroup$
    – Dr. Mathva
    1 hour ago


















$begingroup$
Are you trying to prove the last equality? It's not clear.
$endgroup$
– Victor S.
3 hours ago




$begingroup$
Are you trying to prove the last equality? It's not clear.
$endgroup$
– Victor S.
3 hours ago












$begingroup$
@VictorS. I need to prove that each expression is equal to x + y + z - a
$endgroup$
– questions about math
3 hours ago






$begingroup$
@VictorS. I need to prove that each expression is equal to x + y + z - a
$endgroup$
– questions about math
3 hours ago














$begingroup$
This isn't an answer. However, it could help... $$$$Observe that $$(x+y+z-a)-bigg(x-frac{ayz}{x^2}bigg)=frac{ayz+yx^2+zx^2-ax^2}{x^2}=frac{a(yz-x^2)+x^2(y+z)}{x^2}$$ This equals zero iff $$a(x^2-yz)=x^2(y+z)iff a-frac{ayz}{x^2}=y+z$$ $$iff a·bigg(frac{x^2-yz}{x^2}bigg)=y+ziff a=frac{x^2y+x^2z}{x^2-yz}$$
$endgroup$
– Dr. Mathva
1 hour ago






$begingroup$
This isn't an answer. However, it could help... $$$$Observe that $$(x+y+z-a)-bigg(x-frac{ayz}{x^2}bigg)=frac{ayz+yx^2+zx^2-ax^2}{x^2}=frac{a(yz-x^2)+x^2(y+z)}{x^2}$$ This equals zero iff $$a(x^2-yz)=x^2(y+z)iff a-frac{ayz}{x^2}=y+z$$ $$iff a·bigg(frac{x^2-yz}{x^2}bigg)=y+ziff a=frac{x^2y+x^2z}{x^2-yz}$$
$endgroup$
– Dr. Mathva
1 hour ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

Denote $k=xyz$. Let us consider the equation
$$
t-frac{ak}{t^3}=btag{*}
$$ where $b =x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2},$ and $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a 4-th degree polynomial equation
$$
t^4-b t^3-ak=0.
$$ By Vieta's formula, the fourth root $w$ satisfies
$$
x+y+z+w=b text{ and } xyzw=-ak.
$$ Since $k=xyzne 0$ it follows
$$
w=-a=b-x-y-z,
$$ hence we get
$$
b=x+y+z-a=x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}
$$as wanted.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    $$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ gives
    $$x-y+azleft(frac{x}{y^2}-frac{y}{x^2}right)=0$$ or
    $$1+frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$
    Similarly, $$frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and
    $$frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$
    Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or
    $$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$ or
    $$sum_{cyc}(x^2y^2+x^2yz)=0$$ or
    $$sum_{cyc}z^2(x+y)^2=0,$$ which gives $$x+y=x+z=y+z=0$$ or
    $$x=y=z=0,$$ which is impossible.



    Id est, the given is wrong, which says that
    $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}Rightarrow x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$ is true.






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3123743%2fcan-you-help-me-solve-this-algebra-problem%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Denote $k=xyz$. Let us consider the equation
      $$
      t-frac{ak}{t^3}=btag{*}
      $$       where $b =x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2},$ and $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a 4-th degree polynomial equation
      $$
      t^4-b t^3-ak=0.
      $$       By Vieta's formula, the fourth root $w$ satisfies
      $$
      x+y+z+w=b text{ and } xyzw=-ak.
      $$       Since $k=xyzne 0$ it follows
      $$
      w=-a=b-x-y-z,
      $$       hence we get
      $$
      b=x+y+z-a=x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}
      $$      as wanted.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Denote $k=xyz$. Let us consider the equation
        $$
        t-frac{ak}{t^3}=btag{*}
        $$         where $b =x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2},$ and $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a 4-th degree polynomial equation
        $$
        t^4-b t^3-ak=0.
        $$         By Vieta's formula, the fourth root $w$ satisfies
        $$
        x+y+z+w=b text{ and } xyzw=-ak.
        $$         Since $k=xyzne 0$ it follows
        $$
        w=-a=b-x-y-z,
        $$         hence we get
        $$
        b=x+y+z-a=x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}
        $$        as wanted.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Denote $k=xyz$. Let us consider the equation
          $$
          t-frac{ak}{t^3}=btag{*}
          $$           where $b =x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2},$ and $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a 4-th degree polynomial equation
          $$
          t^4-b t^3-ak=0.
          $$           By Vieta's formula, the fourth root $w$ satisfies
          $$
          x+y+z+w=b text{ and } xyzw=-ak.
          $$           Since $k=xyzne 0$ it follows
          $$
          w=-a=b-x-y-z,
          $$           hence we get
          $$
          b=x+y+z-a=x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}
          $$          as wanted.






          share|cite|improve this answer











          $endgroup$



          Denote $k=xyz$. Let us consider the equation
          $$
          t-frac{ak}{t^3}=btag{*}
          $$           where $b =x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2},$ and $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a 4-th degree polynomial equation
          $$
          t^4-b t^3-ak=0.
          $$           By Vieta's formula, the fourth root $w$ satisfies
          $$
          x+y+z+w=b text{ and } xyzw=-ak.
          $$           Since $k=xyzne 0$ it follows
          $$
          w=-a=b-x-y-z,
          $$           hence we get
          $$
          b=x+y+z-a=x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}
          $$          as wanted.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 15 mins ago

























          answered 49 mins ago









          SongSong

          14.5k1635




          14.5k1635























              1












              $begingroup$

              $$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ gives
              $$x-y+azleft(frac{x}{y^2}-frac{y}{x^2}right)=0$$ or
              $$1+frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$
              Similarly, $$frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and
              $$frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$
              Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or
              $$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$ or
              $$sum_{cyc}(x^2y^2+x^2yz)=0$$ or
              $$sum_{cyc}z^2(x+y)^2=0,$$ which gives $$x+y=x+z=y+z=0$$ or
              $$x=y=z=0,$$ which is impossible.



              Id est, the given is wrong, which says that
              $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}Rightarrow x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$ is true.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ gives
                $$x-y+azleft(frac{x}{y^2}-frac{y}{x^2}right)=0$$ or
                $$1+frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$
                Similarly, $$frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and
                $$frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$
                Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or
                $$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$ or
                $$sum_{cyc}(x^2y^2+x^2yz)=0$$ or
                $$sum_{cyc}z^2(x+y)^2=0,$$ which gives $$x+y=x+z=y+z=0$$ or
                $$x=y=z=0,$$ which is impossible.



                Id est, the given is wrong, which says that
                $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}Rightarrow x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$ is true.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ gives
                  $$x-y+azleft(frac{x}{y^2}-frac{y}{x^2}right)=0$$ or
                  $$1+frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$
                  Similarly, $$frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and
                  $$frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$
                  Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or
                  $$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$ or
                  $$sum_{cyc}(x^2y^2+x^2yz)=0$$ or
                  $$sum_{cyc}z^2(x+y)^2=0,$$ which gives $$x+y=x+z=y+z=0$$ or
                  $$x=y=z=0,$$ which is impossible.



                  Id est, the given is wrong, which says that
                  $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}Rightarrow x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$ is true.






                  share|cite|improve this answer









                  $endgroup$



                  $$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ gives
                  $$x-y+azleft(frac{x}{y^2}-frac{y}{x^2}right)=0$$ or
                  $$1+frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$
                  Similarly, $$frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and
                  $$frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$
                  Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or
                  $$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$ or
                  $$sum_{cyc}(x^2y^2+x^2yz)=0$$ or
                  $$sum_{cyc}z^2(x+y)^2=0,$$ which gives $$x+y=x+z=y+z=0$$ or
                  $$x=y=z=0,$$ which is impossible.



                  Id est, the given is wrong, which says that
                  $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}Rightarrow x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$ is true.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 45 mins ago









                  Michael RozenbergMichael Rozenberg

                  104k1892197




                  104k1892197






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3123743%2fcan-you-help-me-solve-this-algebra-problem%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Discografia di Klaus Schulze Indice Album in studio | Album dal vivo | Singoli | Antologie | Colonne...

                      Image
                      Discografie di artisti tedeschi Voce principale: Klaus Schulze . Discografia di Klaus Schulze Klaus Schulze e Lisa Gerrard durante un concerto Album in studio 56 ↙ Album dal vivo 15 ↙ Raccolte 37 ↙ Singoli 4 ↙ Album video 1 ↙ Colonne sonore 4 ↙ Indice 1 Album in studio 1.1 Solisti 1.1.1 Come Richard Wahnfried 1.2 Collaborazioni 1.2.1 Con Rainer Bloss 1.2.2 Con Rainer Bloss ed Ernst Fuchs 1.2.3 Con Lisa Gerrard 1.2.4 Con Andreas Grosser 1.2.5 Con Pete Namlook 1.2.6 Con Pete Namlook e Bill Laswell 1.2.7 Con Michael Shrieve 1.2.8 Con Solar Moon 1.2.9 Con artisti vari 2 Album dal vivo 3 Singoli 4 Antologie 5 Colonne sonore 6 Videografia 7 Collegamenti musica Album in studio | Solisti | 1972 – Irrlicht 1973 – Cyborg 1974 – Blackdance 1975 – Timewind 1975 – Picture Music 1976 – Moondawn 1977 – Mirage 1978 – "X" 1979 – Dune...
                      Read more

                      Armoriale delle famiglie italiane (Car) Indice Armi | Bibliografia | Menu di navigazioneBlasone...

                      Image
                      Armoriali delle famiglie italiane Blasone Cesenate Famiglie nobili di SiciliaFamiglie nobili di SiciliaTropea MagazineTropea MagazineNobili napoletani Nobili napoletani[1]Nobili napoletaniNobili napoletaniTropea Magazine Nobili napoletaniNobili napoletani Famiglie nobili di SiciliaBlasonario CremoneseFamiglie nobili di Sicilia Archivio Storico CapitolinoRaccolta stemmi TrippiniFamiglie nobili di SiciliaNobili napoletaniArmas de los cavalleros de Veneçia Tropea Magazine Famiglie nobili di Sicilia Nobili napoletaniFamiglie nobili di SiciliaRaccolta stemmi Topoguz Famiglie nobili di SiciliaBlasone CesenateRaccolta stemmi TrippiniArchivio Storico Capitolino Famiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di Sicilia Famiglie nobili di SiciliaFamiglie nobili di SiciliaBlasone Cesenate Raccolta stemmi TrippiniFamiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di Sicilia Blasonario CremoneseBlasonario Cr...
                      Read more

                      Lupi Siderali Indice Storia | Organizzazione | La Tredicesima Compagnia | Aspetto | Membri Importanti...

                      Capitoli degli Space Marine Space MarineImperium della Terrawargame tridimensionaleWarhammer 40.000PrimarcalupiuroImperatoreRune di DissimulazioneAngeli OscuriEresia di HorusStirpe dei MilleOcchio del TerroreCaosTzeentchHorusCodex AstartesSeconda FondazioneFratelli del LupoOcchio del TerroreAlbero della VitareliquieGuardia ImperialeLogan Grimnartruppe d'assaltoImperium dell'UomoveteraniStirpe dei MilleCaosOcchio del TerroreAngeli SanguinariEresia di Horus Lupi Siderali Nome originale Space Wolves Primarca Leman Russ Grido di battaglia Per Russ e per il Padre di Tutti (un feroce ululato) Colori Grigio Altri capitoli I Lupi Siderali sono una delle venti Legioni della Prima Fondazione degli Space Marine che servono l'Imperium della Terra nel wargame tridimensionale Warhammer 40.000. Sono formidabili nel corpo a corpo, i Confratelli desiderano solo la gloria sul campo di battaglia. Indice 1 Storia 1.1 Leman Russ 1.2 La ...
                      Read more