Đáp án:tham khảo
Giải thích các bước giải:
$\text{ Có}$:$x^3(y-x)+z^3(x-y)=y^3(z-x)$
$⇔x^3(y-z)+z^3(x-y)-3y^3(z-x)=0$
$⇔x³y-x³z+z³x-z³y-y³z+y³z=0$
$⇔xy(x²+y²)-xz(x²-z²)-yz(z²+y²)=0$
Có\(\left[ \begin{array}{l}x²+y²≥0\\x²-z²≥0\\x²+y²≥0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}xy=0\\xz=0\\yz=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=0\\y=0\\z=0\end{array} \right.\)
$⇒x+y+z=0⇔x+y=-z$
$\text{Có}$:$x³y³z³-3xyz$
$=(x+y)³+z³-3xy(x+y)-3xyz$
$=-z³+z³-3xy(-z)-xyz$
$=0+3xyz-3xyz$
$=0$
$⇒x³+y³+z³-3xyz=0$
$⇒x³+y³+z³=3xyz$$\text{(đpcm)}$
