$\begin{cases}y+z= -x\\ y^2+z^2=1-x^2 \end{cases}\Leftrightarrow \begin{cases}y^2+2yz+z^2=x^2 \\ y^2+z^2=1-x^2 \end{cases}$
$\Leftrightarrow \begin{cases}y+z=-x \\ yz=\frac{2x^2-1}{2} \end{cases}$
Đặt $a,b$ sao cho $\begin{cases}a=y+z \\ b=yz \end{cases}$
Ta có $x^5+y^5+z^5=x^5+(y^2+z^2)(y^3+z^3)-y^2z^2(y+z)=x^5+a^5-5a^3b+5ab^2$
$= x^5+(-x^5)-5(-x)^3.\frac{2x^2-1}{2}+5(-x)(\frac{2x^2-1}{2})^2=5x^3.\frac{2x^2-1}{2}-5x.\frac{(2x^2-1)^2}{4}$
$=\frac{5}{4}[x^3(2x^2-1)-x(2x^2-1)^2]=\frac{5}{4}(2x^3-x)$