Question

If x+y+z= 6, and x^2+ y^2 + z^2= 18, then find x^3 + y^3 + z^3 - 3xyz ​

Answer 1

Answer:

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

=> (6)^2=18+2(xy+yz+zx)

=> 36-18=2(xy+yz+zx)

=> 18/2=xy+yz+zx

=> xy+yz+zx=9

now,

x^3+y^3+z^3-3xyz

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

=6{18-(xy+yz+zx)}

=6{18-9}

=6.9

=54