I should find $A,B$ and $C$. I know answers but can't figure out how to solve it. Anyone?

We are to find value of $x^4+y^4+z^4$ when $x, y$ and $z$ are real numbers which satisfy the following three equalities: $$x+y+z=3$$ $$x^2+y^2+z^2=9$$ $$xyz=-2$$

Firstly, it follows from the first two equalities that $$xy+yz+zx=A$$

Next, using $$(x^2+y^2+z^2)^2=x^4+y^4+z^4+B[(xy)^2+(yz)^2+(zx)^2],$$ we have $x^4+y^4+z^4=C$.

  • $\begingroup$ There clearly is something wrong with the $B$ line in your question. How is this to be corrected? And is everything after the first line part of the question as was given to you? $\endgroup$ – Rory Daulton Jun 10 '15 at 20:57
  • $\begingroup$ it should be: B{(xy)2+(yz)2+(zx)2} yeah, it's part of a question, here's the image : s013.radikal.ru/i324/1506/56/886b98e3e7dc.png $\endgroup$ – Fiction More Jun 10 '15 at 21:05

Use Newton-Girard relations between elementary symmetric finctions and sums of powers:

Let $p_k=x^k+y^k+z^k$, $\,s_1=x+y+z$, $\,s_2=xy+yz+zx$, $\,s_3=xyz$. We have: \begin{align*} p_1&=s_1& p_2&=s_1p_1-2s_2=s_1^2-2s_2\\ p_3&=s_1p_2-s_2p_1+3s_3&p_4&=s_1p_3-s_2p_2+s_3p_1 \end{align*} We deduce that $s_2=\frac12(s_1^2-p_2)=0$, whence $\,p_3= 21$ and finally $\,p_4=63-6=57$.

  • $\begingroup$ Thank you ! can you tell me what will be the value of B? also, is there a video that explains this formula with examples? thanks again $\endgroup$ – Fiction More Jun 10 '15 at 21:38
  • $\begingroup$ I suppose your formula must be read as $(x^2+y^2+z^2)^2=\dots$? If it is the case, it's well known (multinomial formula) that $B=2$. $\endgroup$ – Bernard Jun 10 '15 at 21:42

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