Question

X + Y + Z is equal to zero, show that x cube + y cube + Z cube = 3 xyz

Answer 1

we know that

x³ + y³ + z³ - xyz = ( 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 friend

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We know that

x^3 + y^3 + z^3 = ( x + y + z ) ( x^2 + y^2 + z^2 -xy -yz - xz ) + 3xyz

Given x + y + z = 0
(substitute in above equation)

x^3 + y^3 + z^3 = 0 ×( x^2 + y^2 + z^2 - xy - yz - zx ) + 3xyz

x^3 + y^3 + z^3 = 3xyz


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Hope it's help you :))