Question

if x+y+z=0 then x³+y³+z³=?​

Answer 1

GIVEN :–

• x + y + z = 0

TO FIND :–

• Value of x³ + y³ + z³ = ?

SOLUTION :–

⇒ x + y + z = 0

⇒ x + y = -z ________eq.(1)

• Cube on both sides –

⇒ (x + y)³ = (-z)³

⇒ (x + y)³ = - z³

• We know that –

➪ (a + b)³ = a³ + b³ + 3ab(a + b)

• So that –

⇒ (x + y)³ = - z³

⇒ x³ + y³ + 3xy(x + y) = - z³

⇒ x³ + y³ + z³ = - 3xy(x + y)

• Using eq.(1) –

⇒ x³ + y³ + z³ = - 3xy(-z)

⇒ x³ + y³ + z³ = 3xyz

Answer 2

Given:

$\rightarrow{\rm{x + y + z = 0}}$

To find:

$\rightarrow{\rm{x^3 + y^3 + z^3}}$

Solution:

This problem is easier to understand. We are going to use the below identity.

$\rightarrow{\rm{a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)}}$

Similarly, we can write as:

$\rightarrow{\rm{x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)}}$

$\rightarrow{\rm{x^3 + y^3 + z^3 - 3xyz = (0)(x^2 + y^2 + z^2 - xy - yz - zx)}}$

We know that any number (or) expression multiplied by 0 is 0. Hence,

$\rightarrow{\rm{x^3 + y^3 + z^3 - 3xyz = 0}}$

$\rightarrow{\boxed{\rm{x^3 + y^3 + z^3 = 3xyz}}}$

Hence, the required answer is 3xyz.