Question

If x+y+z=18 and x square+y square+z square=120 find the value of x cube+y cube+z cube-3xyz

Answer 1

Step-by-step explanation:

x+y+z=18

x^2+y^2+z^2=120

(x+y+z) ^2-2xy-2yz-2zx=120

(18)^2+2(-xy-yz-zx)=120

-xy-yz-zx=(120-324)/2

=-102

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

=18(120-102)

=18×18

=324

hope it will help you