Question

If x+y+z=0, show that x' + y + z = 3xyz
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Answer 1

Answer:

OK actually there is an identity as

a^3 + b^3 + c^3 - 3abc = ( a + b + c ) ( a^2 + b^2 + c^2 - ab - bc - ca)

Step-by-step explanation:

so if a+b+c = 0

then=

In RHS it will be =

0 * ( a^2 + b^2 + c^2 - ab - bc - ca)

= 0

then if will shift the ( -3abc) in LHS to RHS it will be-

a^3 + b^3 + c^3 = 3abc

hope it helps...

if so then please mark me as the brainliest...