Question

$if \: x + y + z = 0 \: prove\: that {x}^{3} + {y}^{3} + {z}^{3} = 3xyz$

Answer 1

hey!!

here is your answer >>>

We know that x^3 + y^3 + z^3 - 3xyz = (x + y + z) (x^2+ y^2 + z^2 - xy - yz - zx)

Now if x + y + z = 0, then

x^3 + y^3 + z^3 - 3xyz = ( 0 ) (x^2 + y^2 + z^2 - xy - yz - zx)  .

  subsituting the value of x + y + z

x^3  + y^3 + z^3 - 3xyz = 0 

 as any thing multiplied by 0 is 0 

x^3  + y^3  + z^3 = 3xyz .      


transporting -3xyz form L.H.S to R.H.S making the term positive

Hope this answer helps!

Answer 2

$x + y + z = 0 = > x + y = - z \\ = > (x + y)^{3} = ( - z )^{3} \\ = > x^{3} + y ^{3} + 3xy \: (x + y) = ( - z) = - {z} \\ = > x^{3} + y ^{3} + 3xy ( - z ) = - {z}^{3} \\ = > x^{3} + y ^{3} - 3xyz = 3xyz$

$Hence,(x + y + z) = 0 = > ( {x}^{3} {y}^{3} {z}^{ 3} ) = 3xyz$