Expand Using the Binomial Theorem (3xy^2-5y)^5

${(3 x y^{2} - 5 y)}^{5}$

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

$\sum_{k = 0}^{5} \frac{5!}{(5 - k)! k!} \cdot {(3 x y^{2})}^{5 - k} \cdot {(- 5 y)}^{k}$

Expand the summation.

$\frac{5!}{(5 - 0)! 0!} \cdot {(3 x y^{2})}^{5 - 0} \cdot {(- 5 y)}^{0} + \frac{5!}{(5 - 1)! 1!} \cdot {(3 x y^{2})}^{5 - 1} \cdot {(- 5 y)}^{1} + \dots + \frac{5!}{(5 - 5)! 5!} \cdot {(3 x y^{2})}^{5 - 5} \cdot {(- 5 y)}^{5}$

Simplify the exponents for each term of the expansion.

$1 \cdot {(3 x y^{2})}^{5} \cdot {(- 5 y)}^{0} + 5 \cdot {(3 x y^{2})}^{4} \cdot {(- 5 y)}^{1} + \dots + 1 \cdot {(3 x y^{2})}^{0} \cdot {(- 5 y)}^{5}$

Simplify the polynomial result.

$243 x^{5} y^{10} - 2025 x^{4} y^{9} + 6750 x^{3} y^{8} - 11250 x^{2} y^{7} + 9375 x y^{6} - 3125 y^{5}$

Expand Using the Binomial Theorem (3xy^2-5y)^5

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