Question

If x+y+z=0 show that x³+y³+z³=3xyz

Answer 1

Hey there!

If
$x + y + z = 0$

$x + y = - z \\ \\ \textbf{cubing \: on \: both \: sides}\\ \\ {(x + y)}^{3} = { - z}^{3} \\ \\ {x}^{3} + {y}^{3} + 3xy(x + y) = { - z}^{3} \\ \\ {x}^{3} + {y}^{3} + 3xy( - z) = { - z}^{3} \\ \\ {x}^{3} + {y}^{3} + {z}^{3} = 3xyz$

Hence proved!

Answer 2

$\bold {\huge{Ello!!}}$

$<b >Here's your answer$

_______________________

Given , x + y + z = 0

To prove , x³ + y³ + z³ = 3xyz

Now ,

x + y + z = 0

x + y = - z ---------------- ( i )

( x + y )³ = ( - z )³ [ ★ Qubing both sides ]

x³ + y³ + 3xy ( x + y ) = - z³

x³ + y³ + z³ + 3xy ( - z ) = 0 [ ★ Putting the value of ( x + y ) = - z ]

x³ + y³ + z³ - 3xyz = 0

x³ + y³ + z³ = 3xyz $\boxed{Hence , Proved}$

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$<b><u><marquee direction> Hope it helps !!$