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

Open in App

Solution

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

Also it is given x + y + z = 0

⇒ $x^{3}$ + $y^{3}$ + $z^{3}$ - 3xyz = (0)( $x^{2}$ + $y^{2}$ + $z^{2}$ - xy - yz -zx)

⇒ $x^{3}$ + $y^{3}$ + $z^{3}$ - 3xyz = 0

⇒ $x^{3}$ + $y^{3}$ + $z^{3}$ = 3xyz

Suggest Corrections

1.6k

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