Question

if x+y+z+=0 , show that x3+y3+z3=3xyz​

Answer 1

Step-by-step explanation:

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

as x+y+z=0

x3+y3+z3-3xyz=0

x3+y3+z3=3xyz

Answer 2

If x+y+z=0

Then

x³+y³+z³- 3xyz=

(x+y+y)[x²+y²+z²-xy-yz-zx]

x³+y³+z³-3cyz=

0[ x²+y²+z²-xy-yz-zx]

x³+y³+z³-3xyz= 0

x³+y³+z³= 3xyz

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