Question

If x y z=0
then prove that
x cube y cube z cube=3xyz

Answer 1

Answer:

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

Hence proved

Solution:

Given that

$x+y+z=0$

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

The above expression can be written as like

$x ^ { 3 } + y ^ { 3 } + z ^ { 3 } - 3 x y z = ( x + y + z ) \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - x y - y z - z x \right)$

Substitute the given value $x+y+z=0$ in the above equation

Therefore we will get,

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

If any value is multiplied with zero than total value will be equal to zero

$\begin{array} { l } { x ^ { 3 } + y ^ { 3 } + z ^ { 3 } - 3 x y z = 0 } \\\\ { x ^ { 3 } + y ^ { 3 } + z ^ { 3 } = 3 x y z } \end{array}$

Hence proved.

Answer 2

Step-by-step explanation:

Given, x3 + y3 + z3 = 3xyz

Therefore, x3 + y3 + z3 - 3xyz =0

This means,

x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 - xy - yz - zx)

Now if x + y + z = 0, then......

x3 + y3 + z3 - 3xyz = ( 0 ) (x2 + y2 + z2 - xy - yz - zx) . . . . . . .... ..... ... .. . . .. . ..... [ subsituting the value of x + y + z ]

x3 + y3 + z3 - 3xyz = 0

x3 + y3 + z3 = 3xyz .