Question

If (x+y+z)=0, then show that x^3+y^3+z^3=3xyz

Answer 1

To Show : x+y+z=0,then x³+y³+z³=3xyz

Identity to be used =

x³+y³+z³-3xyz = (x+y+z)(x²+y²+z²-xy-yz-xz)

Substitute x³+y³+z³=0 in

x³+y³+z³-3xyz = (x+y+z)(x²+y²+z²-xy-yz-xz)

x³+y³+z³-3xyz =0= (x+y+z)(x²+y²+z²-xy-yz-xz)

# x³+y³+z³ - 3xyz = 0

x³+y³+z³ = 3xyz

#Hence Verified .

Answer 2

Given that x + y + z = 0

= > x + y = - z -----: ( 1 )


Cube on both sides ,


= > ( x + y )³ = ( - z )³


$\boxed{ \bold{ \underline{( a + b )^{3}= a^{3} + b^{3} + 3ab( a + b ) }}}$


= > ( x )³ + ( y )³ + 3xy( x + y ) = - z³



Putting the value of ( x + y ) from ( 1 ) ,



= > x³ + y² + 3xy( - z ) = - z³

= > x³ + y³ - 3zxy = -z³

= > x³ + y³ + z³ = 3xyz





Hence, Proved.