Expand Using the Binomial Theorem (a-3b)^6

${(a - 3 b)}^{6}$

Step 1

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

$\sum_{k = 0}^{6} \frac{6!}{(6 - k)! k!} \cdot {(a)}^{6 - k} \cdot {(- 3 b)}^{k}$

Step 2

Expand the summation.

$\frac{6!}{(6 - 0)! 0!} {(a)}^{6 - 0} \cdot {(- 3 b)}^{0} + \frac{6!}{(6 - 1)! 1!} {(a)}^{6 - 1} \cdot {(- 3 b)}^{1} + \dots + \frac{6!}{(6 - 6)! 6!} {(a)}^{6 - 6} \cdot {(- 3 b)}^{6}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(a)}^{6} \cdot {(- 3 b)}^{0} + 6 \cdot {(a)}^{5} \cdot {(- 3 b)}^{1} + \dots + 1 \cdot {(a)}^{0} \cdot {(- 3 b)}^{6}$

Step 4

Simplify the polynomial result.

$a^{6} - 18 a^{5} b + 135 a^{4} b^{2} - 540 a^{3} b^{3} + 1215 a^{2} b^{4} - 1458 a b^{5} + 729 b^{6}$

Expand Using the Binomial Theorem (a-3b)^6

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