What is the Value of \(x^5+y^5+z^5\) Given Initial Polynomial Conditions?

[WMDhamnekar]

$\left\{
\begin{array}{rcl}
x+y+z &=& 1\\
x^2+y^2+z^2 &=& 2\\
x^3+y^3+z^3 &=& 3 \\
x^5+y^5+z^5 &=& ?
\end{array}\right.$How to find out $x^5+y^5+z^5=?$

 

[WMDhamnekar]

$\left\{
\begin{array}{rcl}
x+y+z &=& 1\\
x^2+y^2+z^2 &=& 2\\
x^3+y^3+z^3 &=& 3 \\
x^5+y^5+z^5 &=& ?
\end{array}\right.$How to find out $x^5+y^5+z^5=?$

Hello,

For the readers, viewers, visitors, guests, lurkers and mathematical audience of thehttps://mathhelpboards.com, i am reproducing here,the answer to this question given by math expert on other math and science website on internet.

Let $P_n= x^n + y^n + z^n,$ where n is a positive integer, and $S_1= x+y+z= P_1, S_2= xy + yz + zx, S_3=xyz$

Now we have $P_1= S_1=1$
$P_2=S_1P_1-2S_2= 1- 2S_2=2 \Rightarrow S_2=-\frac12$

$P_3=S_1P_2-S_2P_1+ 3S_3= 2+\frac12 +3S_3=3, \Rightarrow S_3=\frac16$

$P_4=S_1P_3-S_2P_2+S_3P_1= 3 +1 + \frac16 =\frac{25}{6}$

$P_5=S_1P_4 -S_2P_3 + S_3P_2 =\frac{25}{6} +\frac32 +\frac13 = \boxed{6}$ Additional information is available at https://en.wikipedia.org/wiki/Newton's_identities