Answer:

here x+y+z=15

x^2+y^2+z^2=83

(x+y+z)^2=15^2

=> x^2+y^2+z^2+2xy+2yz+2zx=225

=> 83+2(xy+yz+zx)=225

=> 2(xy+yz+zx)=225-83

=> xy+yz+zx=142/2

=> xy+yz+zx=71

now, x^3+y^3+z^3-3xyz

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

=15{(x^2+y^2+z^2)-(xy+yz+zx)}

=15{83-71}

=15*12

=180

Verified answer

Given :

x² + y² + z² = 83

And

x + y + z = 15

To Find :

The value of , x³ + y³ + z³ - 3 x y z

Solution :

∵ ( x + y + z )² = x² + y² + z² + 2 ( x y + y z + z x )

And

x + y + z = 15

So, 83 + 2 ( x y + y z + z x ) = ( 15 )²

Or, 2 ( x y + y z + z x ) = 225 - 83

Or, 2 ( x y + y z + z x ) = 142

∴ ( x y + y z + z x ) = \dfrac{142}{2}

2

142

i.e ( x y + y z + z x ) = 71

Again

∵ x³ + y³ + z³ - 3 x y z = ( x + y + z ) [ ( x² + y² + z² ) - ( x y + y z + z x ) ]

Or, = ( 15 ) × [ 83 - 71 ]

Or, = 15 × 12

i.e = 180

Hence, The value of x³ + y³ + z³ - 3 x y z is 180 . Answer