Mark M. answered 09/30/20
Retired math prof. Very extensive Precalculus tutoring experience.
(x2 + y2)3 = (1)(x2)3(y2)0 + 3(x2)2(y2) + 3(x2)(y2)2 + (1)(x2)0(y2)3
= x6 + 3x4y2 + 3x2y4 + y6
Alex S.
asked 09/30/20Use the Binomial Theorem to expand and simplify the expression (x2 + y2)3
Mark M. answered 09/30/20
Retired math prof. Very extensive Precalculus tutoring experience.
(x2 + y2)3 = (1)(x2)3(y2)0 + 3(x2)2(y2) + 3(x2)(y2)2 + (1)(x2)0(y2)3
= x6 + 3x4y2 + 3x2y4 + y6
(x2 + y2)3 = (x2)3(y2)0 + 3[(x2)2(y2)1] + 3[(x2)1(y2)2] + (x2)0(y2)3
= x6y0 + 3(x4y2)+ 3(x2y4)+ x0y6
= x6(1) + 3x4y2+ 3x2y4+ (1)y6
= x6 + 3x4y2+ 3x2y4+ y6
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