Question

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

Answer 1

Answer

• x³ + y³ + z³ = 3xyz

Given

• x + y + z = 0

To Prove

• x³ + y³ + z³ = 3xyz

Proof

$\rm x+y+z=0\\\\\rightarrow \rm x+y=-z...(1)$

Cubing on both sides , we get ,

$\rightarrow \rm (x+y)^3=(-z)^3\\\\\rightarrow \rm x^3+y^3+3xy(x+y)=-z^3\\\\\rightarrow \rm x^3+y^3+3xy(-z)=-z^3\ \; [From\ (1)]\\\\\rightarrow \rm x^3+y^3-3xyz+z^3=0\\\\\rightarrow \bf \; x^3+y^3+z^3=3xyz$

Hence proved

Answer 2

Given:

• x + y + z = 0

To Prove:

• x³+y³+z³ = 3xyz.

Proof:

Given that x + y + z = 0.

And we are required to prove that x³+y³+z³=3xyz.

Taking the given condition,

=> x + y + z = 0.

=> x + y = -z

Cubing both sides,

=> (x+y)³=(-z)³

=> x³+y³+3xy(x+y) = -z³

Now we already have found that x + y = (-z).

=> x³+y³ + 3xy × (-z) = -z³

=> x³+y³-3xyz = -z³

=> x³+y³+z³=3xyz

Hence proved.