Question

when x=2 and y=(-3)
(x + y)³=x³+y³+3x²y + 3xy²



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Answer 1

Answer:

x=2 & y=(-3)

(x+y)³=x³+y³+3x²y+3xy²

then

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

= 8-27+ 3×4×-3+ 3×2×9

= 8-27 -36 + 54

= -1