Expand Using the Binomial Theorem (x-2/(x^2))^9

${(x - \frac{2}{x^{2}})}^{9}$

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

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

Expand the summation.

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

Simplify the exponents for each term of the expansion.

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

Simplify the polynomial result.

$x^{9} - 18 x^{6} + 144 x^{3} - 672 + \frac{2016}{x^{3}} - \frac{4032}{x^{6}} + \frac{5376}{x^{9}} - \frac{4608}{x^{12}} + \frac{2304}{x^{15}} - \frac{512}{x^{18}}$

Expand Using the Binomial Theorem (x-2/(x^2))^9

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