Question

13. If x + y + z = 0, show that x^3 + y^3 + z^3 = 3xyz​

Answer 1

Given : x + y + z = 0

To prove : x³ + y³ + z³ = 3xyz

Proof : We know that

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

Substituting the value of (x+y+z) = 0

$\implies$x³ + y³ + z³ -3xyz = 0 (x² + y² +z² -xy - yz - zx)

$\implies$x³ + y³ + z³ -3xyz = 0

$\implies$x³ + y³+ z³ = 3xyz

In such types of questions , be clear with your identities and think about in which identity , the given and to prove term come together . This would help you to have the first step towards solving your question.

Some other basic identities :

1. (a+b)² = a² + b² + 2ab
2. (a-b)² = a² + b² -2ab
3. a² - b² = (a+b)(a-b)
4. (a+b+c)² = a² + b² + c² +2ab + 2bc + 2ca
5. (a+b)³ = a³ + b³ + 3a²b + 3ab²
6. (a-b)³ = a³ - b³ -3a²b +3ab²
7. a³ - b³ = (a-b) (a² + b² +ab)

Answer 2

ANSWER:-

we known that

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

puttingx+y+z=0

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

x^3 + y^3 +  z^3 -3xyz=0

then  x^3 + y^3+ z^3=3xyz

therefore proved