Question

If x + y + z = 0 then x³ + y³ + z³ =​

Answer 1

$\boxed{\bf{\red{If \: x + y + z = 0 \: then \: x³ + y³ + z³ = 3xyz}}}$

If x + y + z = 0 then x³ + y³ + z³ = 3xyz

$\pink\dashrightarrow$ x + y + z = 0

$\pink\dashrightarrow$ x + y = - z --( 1 )

$\bold\purple{Cube \: both \:sides}$

$\pink\dashrightarrow$ (x + y)³ = (- z)³

$\quad\bigstar{\underline{\boxed{\green{\sf{Identity \: of \: (x + y)³ = x³ + y³ + 3xy (x + y) }}}}}$

$\pink\dashrightarrow$ x³ + y³ + 3xy (x + y) = (- z)³ ----------- ( 2 )

$\pink\dashrightarrow$ x³ + y³ + 3xy × -z = - z³

$\pink\dashrightarrow$ x³ + y³ - 3xyz = - z³

$\pink\dashrightarrow$ x³ + y³ + z³ = 3xyz

$\bold\red{Hence \: proved}$

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Answer 2

║Answer║

x³ + y³ + z³ = 3xyz

Step-by-step explanation :-

We have,

 x + y + z = 0

⇒ x + y = - z  ____(1)

Taking whole cube on both sides -

(x + y)³ = (-z)³

⇒ x³ + y³ + 3xy(x+y) = (-z)³

⇒ x³ + y³ + z³ = -3xy(x+y) ___(2)

From equation (1)  we have :-

x + y = -z

Now equation (2) becomes :-

x³ + y³ + z³ = -3xy(-z)

⇒ x³ + y³ + z³ = 3xyz

_____..._____

Hope it helps you !!