If x = 2, y = 1, z = -3, then x3 + y3 + z3 - 3xyz equals

Concept

For any three numbers, x3+y3+z3−3xyz factors as (x+y+z)(x2+y2+z2−xy−yz−zx). A key corollary follows directly: whenever x+y+z=0, the entire expression collapses to 0, regardless of the individual values of x, y, and z.

Application

1. Check the sum: x+y+z = 2+1+(-3) = 0.

2. Apply the identity: since the first factor (x+y+z) is 0, the product (x+y+z)(x2+y2+z2−xy−yz−zx) is 0 too, so x3+y3+z3−3xyz = 0.

Cross-check

1. x3 = 23 = 8, y3 = 13 = 1, z3 = (-3)3 = -27, so x3+y3+z3 = -18.

2. xyz = (2)(1)(-3) = -6, so 3xyz = -18.

3. Then -18 - (-18) = 0, matching the result above.

Result

So the expression equals 0.