Question

Answer the question. ​

Answer 1

Answer:

Step-by-step explanation:

$x^{2} /yz$ + $y^2/xz$ + $z^2/xy$ = $x^3 + y^3 + z^3$/xyz -----> (1)

$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - -xy -yz -zx)$

as x + y + z = 0 so $x^3 + y^3 + z^3 - 3xyz= 0$

$x^3 + y^3 + z^3 = 3xyz$

on sub in (1) 3xyz/xyz = 3

Hence proved.