Evaluate x 3+ y 3+ z3- 3xyz when x=-2'y = -1 and z = 3
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Answer:x3+y3+z3-3xyz
substituting x=-2, y=-1, z=3
= (-2)^3 + (-1)^3+ (3)^3-3(-2)(-1)(3)
= -8-1+27–24
= -33+27
=. -6. (ANSWER)
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