If $x + y + z = 0$ , show that $x^{3} + y^{3} + z^{3} = 3 x y z$ .

Open in App

Solution

We know that $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$

Given $x + y + z = 0$

$x^{3} + y^{3} + z^{3} - 3 x y z = (0) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$

$\Rightarrow x^{3} + y^{3} + z^{3} - 3 x y z = 0$

$\Rightarrow x^{3} + y^{3} + z^{3} = 3 x y z$

Hence proved.

Suggest Corrections

Similar questions

x^{3} + y^{3} + z^{3} = 3 x y z .

(a^{3} + b^{3} + c^{3} - 3 a b c) (x^{3} + y^{3} + z^{3} - 3 x y z) = (X^{3} + Y^{3} + Z^{3} - 3 X Y Z)

\frac{x}{a} = \frac{y}{b} = \frac{z}{c}

\frac{x^{3}}{a^{3}} + \frac{y^{3}}{b^{3}} + \frac{z^{3}}{c^{3}} = \frac{3 x y z}{a b c}

x + y + z = 0

(x^{3} + y^{3} + z^{3} - 3 x y z)

x + y + z = 6

x y + y z + z x = 12

x^{3} + y^{3} + z^{3} = 3 x y z

Join BYJU'S Learning Program