Question

if x+y+z=9and xy+yz+zx=23 then the value of x3+y3+z3-3xxyz=?

Answer 1

given
x+y+z=9
xy+yz+zx= 23
x^3+y^3+z^3- xyz =????
so( x+y+z)^2= x^2+y^2+z^2+2 ( xy+yz+zx )
81= x^2+y^2+z^2+2 (23)
81-46 = x^2+y^2+z^2
35=x^2+y ^2+z^2
now
x^3+y^3+z^3-3xyz = (x+y+z) (x^2+y^2+z^2-xy -yz-zx)
= 9× 35 -(23 )
=9× 12
108 is the answer