Expand Using the Binomial Theorem (x^2-1/x)^15

${(x^{2} - \frac{1}{x})}^{15}$

Step 1

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

$\sum_{k = 0}^{15} \frac{15!}{(15 - k)! k!} \cdot {(x^{2})}^{15 - k} \cdot {(- \frac{1}{x})}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x^{2})}^{15} \cdot {(- \frac{1}{x})}^{0} + 15 \cdot {(x^{2})}^{14} \cdot {(- \frac{1}{x})}^{1} + \dots + 1 \cdot {(x^{2})}^{0} \cdot {(- \frac{1}{x})}^{15}$

Step 4

Simplify the polynomial result.

$x^{30} - 15 x^{27} + 105 x^{24} - 455 x^{21} + 1365 x^{18} - 3003 x^{15} + 5005 x^{12} - 6435 x^{9} + 6435 x^{6} - 5005 x^{3} + 3003 - \frac{1365}{x^{3}} + \frac{455}{x^{6}} - \frac{105}{x^{9}} + \frac{15}{x^{12}} - \frac{1}{x^{15}}$

Expand Using the Binomial Theorem (x^2-1/x)^15

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