Question

solve x3+y3+z3-3xyz? when x=2,y=1,z=(-3)

Answer 1

x^3 + y^3 + z^3 -3xyz    [ x = 2  , y = 1  , z = -3  ]

substitute the x , y ,z value in the equation

= (2)^3  +  ( 1 )^3 + (-3)^3 - 3×2×1×-3

= 8 + 1 - 27 + 18

= 8 + 1 + 18 - 27

= 9 + 18 - 27

= 27 - 27

= 0