If x + y + z = 8 and xy +yz +zx = 20, find the value of x^{3} + y^{3}^{ }+ z^{3} −3xyz

#### Solution

n the given problem, we have to find value of x^{3} + y^{3}^{ }+ z^{3} −3xyz

Given x + y + z = 8 , xy +yz +zx = 20

We shall use the identity

`(x+y+z)^2 = x^2 + y^2 + z^2 + 2 (xy + yz +za)`

`(x+y+z)^2 = x^2 + y^2 + z^2 +2 (20)`

`64 = x^2 + y^2 +z^2 + 40`

`64 - 40 = x^2 + y^2 + z^2`

`24 = x^2 + y^2 + z^2`

We know that

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

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

Here substituting `x+y +z = 8,xy +yz + zx = 20,x^2 +y^2 + z^2 = 24 ` we get

`x^3 + y^3 + z^3 -3xyz = 8 [(24 - 20)] `

` = 8 xx 4`

` =32`

Hence the value of x^{3} + y^{3}^{ }+ z^{3} −3xyz is 32.