Expand Using the Binomial Theorem (x-3)^9

${(x - 3)}^{9}$

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

$\sum_{k = 0}^{9} \frac{9!}{(9 - k)! k!} \cdot {(x)}^{9 - k} \cdot {(- 3)}^{k}$

Expand the summation.

$\frac{9!}{(9 - 0)! 0!} {(x)}^{9 - 0} \cdot {(- 3)}^{0} + \frac{9!}{(9 - 1)! 1!} {(x)}^{9 - 1} \cdot (- 3) + \dots + \frac{9!}{(9 - 9)! 9!} {(x)}^{9 - 9} \cdot {(- 3)}^{9}$

Simplify the exponents for each term of the expansion.

$1 \cdot {(x)}^{9} \cdot {(- 3)}^{0} + 9 \cdot {(x)}^{8} \cdot (- 3) + \dots + 1 \cdot {(x)}^{0} \cdot {(- 3)}^{9}$

Simplify the polynomial result.

$x^{9} - 27 x^{8} + 324 x^{7} - 2268 x^{6} + 10206 x^{5} - 30618 x^{4} + 61236 x^{3} - 78732 x^{2} + 59049 x - 19683$

Expand Using the Binomial Theorem (x-3)^9

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