Question

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

Answer 1

${\large{\bold{\bf{\underbrace{Given :-}}}}}$

x + y + z = 0

${\large{\bold{\bf{\underbrace{To \: prove :-}}}}}$

x³ + y³ + z³ = 3xyz

${\large{\bold{\bf{\underbrace{Proof :-}}}}}$

$\pink\dashrightarrow$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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