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Please help me. There are equations: $x+y+z=3, x^2+y^2+z^2=5$ and $x^3+y^3+z^3=7$. The question: what is the result of $x^4+y^4+z^4$?

Ive tried to merge the equation and result in desperado. :( Please explain with simple math as I'm only a junior high school student. Thx a lot

Pebrianto
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    This wiki page may help: http://en.wikipedia.org/wiki/Newton%27s_identities – emeu Nov 08 '13 at 10:18
  • This one also: http://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial – emeu Nov 08 '13 at 10:20
  • Yeah, all people I ask tell me to read about newton sums. But, I dont get it. Please help me.... – Pebrianto Nov 08 '13 at 10:25
  • Same question but solved using Newton's identities. http://math.stackexchange.com/questions/468506/finding-x4-y4-z4-using-geometric-series – joejacobz Nov 08 '13 at 10:36

3 Answers3

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use this

since $$x^2+y^2+z^2=(x+y+z)^2-2xy-2yz-2xz\Longrightarrow xy+yz+xz=2$$ $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=(x+y+z)^3-3(xy+yz+xz)(x+y+z)$$ so $$7-3xyz=27-18\Longrightarrow xyz=-\dfrac{2}{3}$$ use $$x^4+y^4+z^4=(x+y+z)(x^3+y^3+z^3)-(xy+yz+xz)(x^2+y^2+z^2)+xyz(x+y+z)$$

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solution 2: since $$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2x^2y^2-2y^2z^2-2x^2z^2$$ and $$(xy+yz+xz)^2=x^2y^2+x^2z^2+y^2z^2+2xyz(x+y+z)$$ since $$xy+yz+xz=2,xyz=-\dfrac{2}{3}$$ so $$x^2y^2+y^2z^2+x^2z^2=8$$ so $$x^4+y^4+z^4=5^2-2\cdot 8=9$$

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We have in general if $$ s_1:=x+y+z\textrm{, }s_2:=x^2+y^2+z^2\textrm{, }s_3:=x^3+y^3+z^3 $$ and $$ s_4:=x^4+y^4+z^4 $$ and if $$ \sigma_1=x+y+z\textrm{, }\sigma_2=xy+yz+zx\textrm{, }\sigma_3=xyz\tag 1 $$ Then $$ s_1=\sigma_1\textrm{, }s_2=\sigma_1^2-2\sigma_2\textrm{, }s_3=\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3\textrm{, }s_4=\sigma_1^4-4\sigma_1^2\sigma_2+2\sigma_2^2+4\sigma_1 \sigma_3\tag 2 $$ Hence if $s_1=3,s_2=5,s_3=7$, then $$ \sigma_1=s_1=3\textrm{, }\sigma_2=2\textrm{, }\sigma_3=-2/3 $$ Hence $$ s_4=x^4+y^4+z^4=9 $$ For to find $x^5+y^5+z^5$, we use $$ (x+y+z)^5=x^5+y^5+z^5+20xyz(x^2+y^2+z^2)+30xyz(xy+yz+zx)+5 S, $$ where $$ S=\sum_{cyc}\left(x^4y+xy^4+2x^3y^2+2x^2y^3\right) =\sum_{cyc}xy\left(x^3+y^3\right)+2x^2y^2z^2\sum_{cyc}\frac{x+y}{z^2}= $$ $$ =\sum_{cyc}xy\left(x^3+y^3+z^3-z^3\right)+2\sigma_3^2\sum_{cyc}\frac{s_1-z}{z^2}= $$ $$ =s_3\sum_{cyc}xy-\sigma_3s_2+2\sigma_3^2s_1\sum_{cyc}x^{-2}-2\sigma_3^2\sum_{cyc}x^{-1}= $$ $$ =s_3\sigma_2-\sigma_3s_2+2\sigma_3^2s_1\frac{-\sigma_2^2+2\sigma_1\sigma_2}{\sigma_3^2}-2\sigma_3^2\frac{\sigma_2}{\sigma_3}. $$ Since $$ \sum_{cyc}\frac{1}{x^2}=\frac{2\sigma_1\sigma_2-2\sigma_2^2}{\sigma_3^2}\textrm{, }\sum_{cyc}\frac{1}{x}=\frac{\sigma_2}{\sigma_3}. $$ Hence finaly in general $$ s_5=x^5+y^5+z^5=\sigma_1^5-5\sigma_1^3\sigma_2+25\sigma_1\sigma_2^2-5\sigma_2\sigma_3-5\sigma_1^2(4\sigma_2+3\sigma_3).\tag 3 $$ and as application $$ x^5+y^5+z^5=29/3. $$

Nikos Bagis
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