Expand Using the Binomial Theorem (3u+v^2)^6

${(3 u + v^{2})}^{6}$

Step 1

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

$\sum_{k = 0}^{6} \frac{6!}{(6 - k)! k!} \cdot {(3 u)}^{6 - k} \cdot {(v^{2})}^{k}$

Step 2

Expand the summation.

$\frac{6!}{(6 - 0)! 0!} \cdot {(3 u)}^{6 - 0} \cdot {(v^{2})}^{0} + \frac{6!}{(6 - 1)! 1!} \cdot {(3 u)}^{6 - 1} \cdot {(v^{2})}^{1} + \dots + \frac{6!}{(6 - 6)! 6!} \cdot {(3 u)}^{6 - 6} \cdot {(v^{2})}^{6}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(3 u)}^{6} \cdot {(v^{2})}^{0} + 6 \cdot {(3 u)}^{5} \cdot {(v^{2})}^{1} + \dots + 1 \cdot {(3 u)}^{0} \cdot {(v^{2})}^{6}$

Step 4

Simplify the polynomial result.

$729 u^{6} + 1458 u^{5} v^{2} + 1215 u^{4} v^{4} + 540 u^{3} v^{6} + 135 u^{2} v^{8} + 18 u v^{10} + v^{12}$

Expand Using the Binomial Theorem (3u+v^2)^6

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