Expand Using the Binomial Theorem (x+1/(x^2))^6

${(x + \frac{1}{x^{2}})}^{6}$

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

$\sum_{k = 0}^{6} \frac{6!}{(6 - k)! k!} \cdot {(x)}^{6 - k} \cdot {(\frac{1}{x^{2}})}^{k}$

Expand the summation.

$\frac{6!}{(6 - 0)! 0!} {(x)}^{6 - 0} \cdot {(\frac{1}{x^{2}})}^{0} + \frac{6!}{(6 - 1)! 1!} {(x)}^{6 - 1} \cdot {(\frac{1}{x^{2}})}^{1} + \dots + \frac{6!}{(6 - 6)! 6!} {(x)}^{6 - 6} \cdot {(\frac{1}{x^{2}})}^{6}$

Simplify the exponents for each term of the expansion.

$1 \cdot {(x)}^{6} \cdot {(\frac{1}{x^{2}})}^{0} + 6 \cdot {(x)}^{5} \cdot {(\frac{1}{x^{2}})}^{1} + \dots + 1 \cdot {(x)}^{0} \cdot {(\frac{1}{x^{2}})}^{6}$

Simplify the polynomial result.

$x^{6} + 6 x^{3} + 15 + \frac{20}{x^{3}} + \frac{15}{x^{6}} + \frac{6}{x^{9}} + \frac{1}{x^{12}}$

Expand Using the Binomial Theorem (x+1/(x^2))^6

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