Expand Using the Binomial Theorem (1/(x^3)-2x)^16

${(\frac{1}{x^{3}} - 2 x)}^{16}$

Step 1

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

$\sum_{k = 0}^{16} \frac{16!}{(16 - k)! k!} \cdot {(\frac{1}{x^{3}})}^{16 - k} \cdot {(- 2 x)}^{k}$

Step 2

Expand the summation.

$\frac{16!}{(16 - 0)! 0!} \cdot {(\frac{1}{x^{3}})}^{16 - 0} \cdot {(- 2 x)}^{0} + \frac{16!}{(16 - 1)! 1!} \cdot {(\frac{1}{x^{3}})}^{16 - 1} \cdot {(- 2 x)}^{1} + \dots + \frac{16!}{(16 - 16)! 16!} \cdot {(\frac{1}{x^{3}})}^{16 - 16} \cdot {(- 2 x)}^{16}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(\frac{1}{x^{3}})}^{16} \cdot {(- 2 x)}^{0} + 16 \cdot {(\frac{1}{x^{3}})}^{15} \cdot {(- 2 x)}^{1} + \dots + 1 \cdot {(\frac{1}{x^{3}})}^{0} \cdot {(- 2 x)}^{16}$

Step 4

Simplify the polynomial result.

$\frac{1}{x^{48}} - \frac{32}{x^{44}} + \frac{480}{x^{40}} - \frac{4480}{x^{36}} + \frac{29120}{x^{32}} - \frac{139776}{x^{28}} + \frac{512512}{x^{24}} - \frac{1464320}{x^{20}} + \frac{3294720}{x^{16}} - \frac{5857280}{x^{12}} + \frac{8200192}{x^{8}} - \frac{8945664}{x^{4}} + 7454720 - 4587520 x^{4} + 1966080 x^{8} - 524288 x^{12} + 65536 x^{16}$

Expand Using the Binomial Theorem (1/(x^3)-2x)^16

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