Question

If x+y+z=9 and x^2+y^2+z^2=35, find the value of x^3 + y^3+z^3-3xyz​

Answer 1

Answer is above mentioned.

Answer 2

x+y+z=9

Squaring both sides, we get

x^2+y^2+z^2+2xy+2yz+2xz= 81

35+2(xy+yz+xz)= 81

xy+yz+xz= 23

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

=(9)(35-23)

=9*12=108