Question

If x+y+z=9 and xy+yz+zx=23 then find the values of (x^3+y^3+z^3-3xyz)?​

Answer 1

Answer:

x3 + y3 + z3 - 3xyz)

= (x + y + z) (x2 + y2 + z2 - xy - yz - zx)

= (x + y + z) [(x + y + z)2 - 3(xy + yz + zx)]

= 9 x (81 - 3 x 23)

= (9 x 12)

= 108