A3+b3
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(a+b)3=a3+b3+3ab(a+b) a3+b3=(a+b)3-3ab(a+b) a3+b3=(a+b)(a2-ab+b2)
a3*b3 = a3b3
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(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a -b) · (a2+ab+b2) = (a3+a2b+ab2) - (a2b+ab2+b3) = a3 -b3 (a+b) · (a2 -ab+b2) = (a3 -a2b+ab2) +(a2b -ab2+b3) = a3+b3 More generally: (a ∓ b) · (an-1 ±an-2b +an-3b2 ±an-4b3 +±...+a(±b)n-2 +(±b)n-1) = an ± bn. The mixed terms cancel out themselves.