Expand Using the Binomial Theorem (x^2+1/x)^9

${(x^{2} + \frac{1}{x})}^{9}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify the polynomial result.

$x^{18} + 9 x^{15} + 36 x^{12} + 84 x^{9} + 126 x^{6} + 126 x^{3} + 84 + \frac{36}{x^{3}} + \frac{9}{x^{6}} + \frac{1}{x^{9}}$

Expand Using the Binomial Theorem (x^2+1/x)^9

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