Question

if x=98, y=99, z=100. find x³+y³+z³-3xyz​

Answer 1

a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

x³+y³+z³-3xyz = (98+99+100) (98^2+99^2+100^2−98×99−99×100−100×98)

=297*3

=891