Question

If x+y+z=0 show that x cube+y cube+z cube=3xyz

Answer 1

Answer:

Step-by-step explanation:

Using formula x3+y3+z3-3xyz=(X+Y+Z)(X2+Y2+Z2-xy-yz-zx)....(1)

Given x+y+z=o

Then pitiing that value in (1)

We get. .

X3+y3+z3-3xyz=(0)(x2+y2+z2-xy-yz-zx)

X3+y3+z3-3xyz=0

X3+y3+z3=3xyz

Hence proved.

Here we go✌

Answer 2

HELLO DEAR FRIEND

GIVEN

$\large{x + y + z = 0}$

WE KNOW THAT

$\large{ {x}^{3} + {y}^{3} + {z}^{3} - 3xyz =}$

$\large{(x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - xz - yz) }$

if

$\large{x + y + z = 0}$

Then

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

so.,

$\large{ {x}^{3} + {y}^{3} + {z}^{3} = 3xyz}$

HOPE IT HELPS YOU DEAR FRIEND THANKS