Expand Using the Binomial Theorem (2xy^3+y)^7

${(2 x y^{3} + y)}^{7}$

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{7} \frac{7!}{(7 - k)! k!} \cdot {(2 x y^{3})}^{7 - k} \cdot {(y)}^{k}$

Expand the summation.

$\frac{7!}{(7 - 0)! 0!} \cdot {(2 x y^{3})}^{7 - 0} \cdot {(y)}^{0} + \frac{7!}{(7 - 1)! 1!} \cdot {(2 x y^{3})}^{7 - 1} \cdot {(y)}^{1} + \dots + \frac{7!}{(7 - 7)! 7!} \cdot {(2 x y^{3})}^{7 - 7} \cdot {(y)}^{7}$

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 x y^{3})}^{7} \cdot {(y)}^{0} + 7 \cdot {(2 x y^{3})}^{6} \cdot {(y)}^{1} + \dots + 1 \cdot {(2 x y^{3})}^{0} \cdot {(y)}^{7}$

Simplify the polynomial result.

$128 x^{7} y^{21} + 448 x^{6} y^{19} + 672 x^{5} y^{17} + 560 x^{4} y^{15} + 280 x^{3} y^{13} + 84 x^{2} y^{11} + 14 x y^{9} + y^{7}$

Expand Using the Binomial Theorem (2xy^3+y)^7

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