Question

x+ y + Z is equal to zero show that xcube + y cube + Z cube is equal to 3 x y z

Answer 1

Heya mate, Here is ur answer

x+y+z=0

x+y=-z

Cubing both the sides

(x+y)^3=-z^3

${x}^{3} + y {}^{3} + 3xy(x + y) = - {z}^{3}$

x+y=-z

${x}^{3} + y {}^{3} + 3xy \times - z = - z {}^{3}$

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

Transposing the terms

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


Hence, proved

Warm regards

@Laughterqueen

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Answer 2

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$\underline\bold{\huge{ANSWER \: :}}$

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Given, x+y+z = 0.
=> x+y = -z.

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Now, Cubing both sides, we get :

(x+y)³=(-z)³

=> x³+y³+3xy(x+y) = (-z)³

=> x³+y³+3xy (-z) = (-z)³
[Putting the value of (x+y) = (-z)]

=> x³+y³+z³ = 3xyz [PROVED]

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