Question

If x + y + z = 8 and xy +yz +zx = 20, find the value of x³ + y³ + z³ −3xyz

Answer 1

n the given problem, we have to find value of  x³ + y³ + z³ −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  x³ + y³ + z³ −3xyz is 32.