Question

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

Answer 1

we know that

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

Putting x + y + z = 0 on R.H.S ., we get

x³ + y³ + z³ - 3xyz = 0

→ x³ + y³ + z³ = 3xyz

Answer 2

Hello dear friend ...
Ur answer is .

SOL.
$x + y + z = 0 \\ {(x + y + z)}^{3} = {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = 0 \\ \\ = {x}^{3} + {y}^{3} + {z}^{3} = 3xyz \: \: \: \: proved$
Here, proved

Hope it's helps you.
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