<s>[INST] Find value of $x^{5} + y^{5} + z^{5}$ given the values of $x + y + z$, $x^{2} + y^{2} + z^{2}$ and $x^{3} + y^{3} + z^{3}$ If$$x+y+z=1$$
$$x^2+y^2+z^2=2$$
$$x^3+y^3+z^3=3$$
Then find the value of $$x^5+y^5+z^5$$
Is there any simple way to solve this problem ? I have tried all my tricks tried to multiply two equations , substitute $z=1-x-y$ , but things got messy nothing seems to work out .
 [/INST] This form of equations can be solved systematically using Newton's identities.
The problem at hand is a special case of $3$ variables. Let $e_1,e_2,e_3$ be the three elementary symmetric polynomial associated with $x,y,z$:
$$\begin{align}
e_1 &= x + y + z\\
e_2 &= xy+ yz + zx\\
e_3 &= xyz
\end{align}$$
This is equivalent to $\displaystyle\;(\lambda - x)(\lambda -y)(\lambda -z) = \lambda^3 - e_1\lambda^2 + e_2\lambda - e_3$.
For any integer $k > 0$, let $p_k = x^k + y^k + z^k$.
For 3 variables, the Newton's identities are:
$$\begin{align}
p_1 &= e_1\\
p_2 &= e_1 p_1 - 2 e_2\\
p_3 &= e_1 p_2 - e_2 p_1 + 3e_3\\
p_n &= e_1 p_{n-1} - e_2 p_{n-2} + e_3 p_{n-3} \quad\text{ for }\; n > 3
\end{align}$$
Given the known values of $(p_1,p_2,p_3) = (1,2,3)$, we have
$$\begin{align}
e_1 &= p_1 = 1;\\
e_2 &= -\frac12(p_2 - e_1p_1) = -\frac12\left(2 - 1^2\right) = -\frac12\\
e_3 &= \frac13(p_3 - e_1p_2 + e_2p_1) = \frac13\left(3 - 2 - \frac12\right) = \frac16
\end{align}$$
Substitute this into formula of $p_4$ and $p_5$, we have
$$\begin{align}
p_4 &= e_1 p_3 - e_2 p_2 + e_3 p_1 = 3 + \frac12\cdot 2 + \frac16\cdot 1 = \frac{25}{6}\\
p_5 &= e_1 p_4 - e_2 p_3 + e_3 p_2 = \frac{25}{6} + \frac12\cdot 3 + \frac16\cdot 2 = 6
\end{align}$$
The answer is $x^5 + y^5 + z^5 = 6$.
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