Expand Using the Binomial Theorem (1/2x^3y^2-4y)^9

${(\frac{1}{2} x^{3} y^{2} - 4 y)}^{9}$

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

$\sum_{k = 0}^{9} \frac{9!}{(9 - k)! k!} \cdot {(\frac{1}{2} \cdot (x^{3} y^{2}))}^{9 - k} \cdot {(- 4 y)}^{k}$

Expand the summation.

$\frac{9!}{(9 - 0)! 0!} \cdot {(\frac{1}{2} \cdot (x^{3} y^{2}))}^{9 - 0} \cdot {(- 4 y)}^{0} + \frac{9!}{(9 - 1)! 1!} \cdot {(\frac{1}{2} \cdot (x^{3} y^{2}))}^{9 - 1} \cdot {(- 4 y)}^{1} + \dots + \frac{9!}{(9 - 9)! 9!} \cdot {(\frac{1}{2} \cdot (x^{3} y^{2}))}^{9 - 9} \cdot {(- 4 y)}^{9}$

Simplify the exponents for each term of the expansion.

$1 \cdot {(\frac{1}{2} \cdot (x^{3} y^{2}))}^{9} \cdot {(- 4 y)}^{0} + 9 \cdot {(\frac{1}{2} \cdot (x^{3} y^{2}))}^{8} \cdot {(- 4 y)}^{1} + \dots + 1 \cdot {(\frac{1}{2} \cdot (x^{3} y^{2}))}^{0} \cdot {(- 4 y)}^{9}$

Simplify the polynomial result.

$\frac{x^{27} y^{18}}{512} - \frac{9 x^{24} y^{17}}{64} + \frac{9 x^{21} y^{16}}{2} - 84 x^{18} y^{15} + 1008 x^{15} y^{14} - 8064 x^{12} y^{13} + 43008 x^{9} y^{12} - 147456 x^{6} y^{11} + 294912 x^{3} y^{10} - 262144 y^{9}$

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