Question

If x+y+z = 4, xy + y2 + zx = 1, find the value of x3+y3+z3-3xyz.​

Answer 1

x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)

=4(x²+y²+z²-1)

=4{(x+y+z)²-2xy-2yz-2zx-1)}

=4(16-2-1)

=52