System of 3 equations problem












1















If
$$x+y+z=1\x^2+y^2+z^2={3over2}\x^3+y^3+z^3=1$$
Then how much is
$$x^4+y^4+z^4$$




So what I'm pretty sure about, this shouldn't be solved for $x$,$y$, and $z$, but I should try to make $x^4+y^4+z^4$ from these equations.



I did this:



$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2$



And from here I can see:



$x^4+y^4+z^4=1-2(x^2y^2+x^2z^2+y^2z^2)$



But I'm missing $(x^2y^2+x^2z^2+y^2z^2)$



I tried finding $(x+y+z)^2$ and other combinations of squaring, tried multiplying some of these, but still couldn't find what I needed.



I'm in high school so try and do this for my level of knowledge, with at least a similar way to mine






algebra-precalculus systems-of-equations







share|cite|improve this question





























    1















    If
    $$x+y+z=1\x^2+y^2+z^2={3over2}\x^3+y^3+z^3=1$$
    Then how much is
    $$x^4+y^4+z^4$$




    So what I'm pretty sure about, this shouldn't be solved for $x$,$y$, and $z$, but I should try to make $x^4+y^4+z^4$ from these equations.



    I did this:



    $(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2$



    And from here I can see:



    $x^4+y^4+z^4=1-2(x^2y^2+x^2z^2+y^2z^2)$



    But I'm missing $(x^2y^2+x^2z^2+y^2z^2)$



    I tried finding $(x+y+z)^2$ and other combinations of squaring, tried multiplying some of these, but still couldn't find what I needed.



    I'm in high school so try and do this for my level of knowledge, with at least a similar way to mine






    algebra-precalculus systems-of-equations







    share|cite|improve this question



























      1












      1








      1


      1






      If
      $$x+y+z=1\x^2+y^2+z^2={3over2}\x^3+y^3+z^3=1$$
      Then how much is
      $$x^4+y^4+z^4$$




      So what I'm pretty sure about, this shouldn't be solved for $x$,$y$, and $z$, but I should try to make $x^4+y^4+z^4$ from these equations.



      I did this:



      $(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2$



      And from here I can see:



      $x^4+y^4+z^4=1-2(x^2y^2+x^2z^2+y^2z^2)$



      But I'm missing $(x^2y^2+x^2z^2+y^2z^2)$



      I tried finding $(x+y+z)^2$ and other combinations of squaring, tried multiplying some of these, but still couldn't find what I needed.



      I'm in high school so try and do this for my level of knowledge, with at least a similar way to mine






      algebra-precalculus systems-of-equations







      share|cite|improve this question
















      If
      $$x+y+z=1\x^2+y^2+z^2={3over2}\x^3+y^3+z^3=1$$
      Then how much is
      $$x^4+y^4+z^4$$




      So what I'm pretty sure about, this shouldn't be solved for $x$,$y$, and $z$, but I should try to make $x^4+y^4+z^4$ from these equations.



      I did this:



      $(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2$



      And from here I can see:



      $x^4+y^4+z^4=1-2(x^2y^2+x^2z^2+y^2z^2)$



      But I'm missing $(x^2y^2+x^2z^2+y^2z^2)$



      I tried finding $(x+y+z)^2$ and other combinations of squaring, tried multiplying some of these, but still couldn't find what I needed.



      I'm in high school so try and do this for my level of knowledge, with at least a similar way to mine






      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 Nov 27 at 11:30

























      asked Nov 27 at 11:18









      Aleksa

      33612




      33612






















          2 Answers
          2






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          3














          Hint:



          $$(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$$



          $$(x+y+z)^2=?$$



          $$x^3+y^3+z^3-3xyz=(x+y+z)((x^2+y^2+z^2-(xy+yz+zx))$$






          share|cite|improve this answer





















          • See also: en.wikipedia.org/wiki/Newton%27s_identities
            – lab bhattacharjee
            Nov 27 at 11:35



















          0














          Alternativley, working with trinomials is cumbersome, so make them binomials:
          $$begin{cases}x+y+z=1\x^2+y^2+z^2={3over2}\x^3+y^3+z^3=1end{cases} Rightarrow begin{cases}x+y=1-z\x^2+y^2={3over2}-z^2\x^3+y^3=1-z^3end{cases}.$$
          $(1)^2-(2)$:
          $$2xy=2z^2-2z-frac12$$
          $(1)^3-(1)$:
          $$3xy(x+y)=3z(z-1) Rightarrow 3xy(1-z)=3z(z-1) Rightarrow xy=-z$$
          So:
          $$-2z=2z^2-2z-frac12 Rightarrow z=pm frac12.$$
          $(1)^4$:
          $$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4 Rightarrow \
          x^4+y^4=(x+y)^4-4xy(x^2+y^2)-6(xy)^2 Rightarrow \
          x^4+y^4+z^4=(1-z)^4-4(-z)(frac32-z^2)-6(-z)^2+z^4.$$
          Plugging $z=frac12$:
          $$x^4+y^4+z^4=frac1{16}+frac52-frac32+frac1{16}=frac98.$$
          Plugging $z=-frac12$:
          $$x^4+y^4+z^4=frac{81}{16}-frac52-frac32+frac1{16}=frac98.$$






          share|cite|improve this answer





















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            2 Answers
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            2 Answers
            2






            active

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            active

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            active

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            3














            Hint:



            $$(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$$



            $$(x+y+z)^2=?$$



            $$x^3+y^3+z^3-3xyz=(x+y+z)((x^2+y^2+z^2-(xy+yz+zx))$$






            share|cite|improve this answer





















            • See also: en.wikipedia.org/wiki/Newton%27s_identities
              – lab bhattacharjee
              Nov 27 at 11:35
















            3














            Hint:



            $$(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$$



            $$(x+y+z)^2=?$$



            $$x^3+y^3+z^3-3xyz=(x+y+z)((x^2+y^2+z^2-(xy+yz+zx))$$






            share|cite|improve this answer





















            • See also: en.wikipedia.org/wiki/Newton%27s_identities
              – lab bhattacharjee
              Nov 27 at 11:35














            3












            3








            3






            Hint:



            $$(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$$



            $$(x+y+z)^2=?$$



            $$x^3+y^3+z^3-3xyz=(x+y+z)((x^2+y^2+z^2-(xy+yz+zx))$$






            share|cite|improve this answer












            Hint:



            $$(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$$



            $$(x+y+z)^2=?$$



            $$x^3+y^3+z^3-3xyz=(x+y+z)((x^2+y^2+z^2-(xy+yz+zx))$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 27 at 11:32









            lab bhattacharjee

            223k15156274




            223k15156274












            • See also: en.wikipedia.org/wiki/Newton%27s_identities
              – lab bhattacharjee
              Nov 27 at 11:35


















            • See also: en.wikipedia.org/wiki/Newton%27s_identities
              – lab bhattacharjee
              Nov 27 at 11:35
















            See also: en.wikipedia.org/wiki/Newton%27s_identities
            – lab bhattacharjee
            Nov 27 at 11:35




            See also: en.wikipedia.org/wiki/Newton%27s_identities
            – lab bhattacharjee
            Nov 27 at 11:35











            0














            Alternativley, working with trinomials is cumbersome, so make them binomials:
            $$begin{cases}x+y+z=1\x^2+y^2+z^2={3over2}\x^3+y^3+z^3=1end{cases} Rightarrow begin{cases}x+y=1-z\x^2+y^2={3over2}-z^2\x^3+y^3=1-z^3end{cases}.$$
            $(1)^2-(2)$:
            $$2xy=2z^2-2z-frac12$$
            $(1)^3-(1)$:
            $$3xy(x+y)=3z(z-1) Rightarrow 3xy(1-z)=3z(z-1) Rightarrow xy=-z$$
            So:
            $$-2z=2z^2-2z-frac12 Rightarrow z=pm frac12.$$
            $(1)^4$:
            $$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4 Rightarrow \
            x^4+y^4=(x+y)^4-4xy(x^2+y^2)-6(xy)^2 Rightarrow \
            x^4+y^4+z^4=(1-z)^4-4(-z)(frac32-z^2)-6(-z)^2+z^4.$$
            Plugging $z=frac12$:
            $$x^4+y^4+z^4=frac1{16}+frac52-frac32+frac1{16}=frac98.$$
            Plugging $z=-frac12$:
            $$x^4+y^4+z^4=frac{81}{16}-frac52-frac32+frac1{16}=frac98.$$






            share|cite|improve this answer


























              0














              Alternativley, working with trinomials is cumbersome, so make them binomials:
              $$begin{cases}x+y+z=1\x^2+y^2+z^2={3over2}\x^3+y^3+z^3=1end{cases} Rightarrow begin{cases}x+y=1-z\x^2+y^2={3over2}-z^2\x^3+y^3=1-z^3end{cases}.$$
              $(1)^2-(2)$:
              $$2xy=2z^2-2z-frac12$$
              $(1)^3-(1)$:
              $$3xy(x+y)=3z(z-1) Rightarrow 3xy(1-z)=3z(z-1) Rightarrow xy=-z$$
              So:
              $$-2z=2z^2-2z-frac12 Rightarrow z=pm frac12.$$
              $(1)^4$:
              $$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4 Rightarrow \
              x^4+y^4=(x+y)^4-4xy(x^2+y^2)-6(xy)^2 Rightarrow \
              x^4+y^4+z^4=(1-z)^4-4(-z)(frac32-z^2)-6(-z)^2+z^4.$$
              Plugging $z=frac12$:
              $$x^4+y^4+z^4=frac1{16}+frac52-frac32+frac1{16}=frac98.$$
              Plugging $z=-frac12$:
              $$x^4+y^4+z^4=frac{81}{16}-frac52-frac32+frac1{16}=frac98.$$






              share|cite|improve this answer
























                0












                0








                0






                Alternativley, working with trinomials is cumbersome, so make them binomials:
                $$begin{cases}x+y+z=1\x^2+y^2+z^2={3over2}\x^3+y^3+z^3=1end{cases} Rightarrow begin{cases}x+y=1-z\x^2+y^2={3over2}-z^2\x^3+y^3=1-z^3end{cases}.$$
                $(1)^2-(2)$:
                $$2xy=2z^2-2z-frac12$$
                $(1)^3-(1)$:
                $$3xy(x+y)=3z(z-1) Rightarrow 3xy(1-z)=3z(z-1) Rightarrow xy=-z$$
                So:
                $$-2z=2z^2-2z-frac12 Rightarrow z=pm frac12.$$
                $(1)^4$:
                $$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4 Rightarrow \
                x^4+y^4=(x+y)^4-4xy(x^2+y^2)-6(xy)^2 Rightarrow \
                x^4+y^4+z^4=(1-z)^4-4(-z)(frac32-z^2)-6(-z)^2+z^4.$$
                Plugging $z=frac12$:
                $$x^4+y^4+z^4=frac1{16}+frac52-frac32+frac1{16}=frac98.$$
                Plugging $z=-frac12$:
                $$x^4+y^4+z^4=frac{81}{16}-frac52-frac32+frac1{16}=frac98.$$






                share|cite|improve this answer












                Alternativley, working with trinomials is cumbersome, so make them binomials:
                $$begin{cases}x+y+z=1\x^2+y^2+z^2={3over2}\x^3+y^3+z^3=1end{cases} Rightarrow begin{cases}x+y=1-z\x^2+y^2={3over2}-z^2\x^3+y^3=1-z^3end{cases}.$$
                $(1)^2-(2)$:
                $$2xy=2z^2-2z-frac12$$
                $(1)^3-(1)$:
                $$3xy(x+y)=3z(z-1) Rightarrow 3xy(1-z)=3z(z-1) Rightarrow xy=-z$$
                So:
                $$-2z=2z^2-2z-frac12 Rightarrow z=pm frac12.$$
                $(1)^4$:
                $$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4 Rightarrow \
                x^4+y^4=(x+y)^4-4xy(x^2+y^2)-6(xy)^2 Rightarrow \
                x^4+y^4+z^4=(1-z)^4-4(-z)(frac32-z^2)-6(-z)^2+z^4.$$
                Plugging $z=frac12$:
                $$x^4+y^4+z^4=frac1{16}+frac52-frac32+frac1{16}=frac98.$$
                Plugging $z=-frac12$:
                $$x^4+y^4+z^4=frac{81}{16}-frac52-frac32+frac1{16}=frac98.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 27 at 14:36









                farruhota

                19.2k2736




                19.2k2736






























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