Expand Using the Binomial Theorem (3a+4b)^8

${(3 a + 4 b)}^{8}$

Step 1

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

$\sum_{k = 0}^{8} \frac{8!}{(8 - k)! k!} \cdot {(3 a)}^{8 - k} \cdot {(4 b)}^{k}$

Step 2

Expand the summation.

$\frac{8!}{(8 - 0)! 0!} \cdot {(3 a)}^{8 - 0} \cdot {(4 b)}^{0} + \frac{8!}{(8 - 1)! 1!} \cdot {(3 a)}^{8 - 1} \cdot {(4 b)}^{1} + \dots + \frac{8!}{(8 - 8)! 8!} \cdot {(3 a)}^{8 - 8} \cdot {(4 b)}^{8}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(3 a)}^{8} \cdot {(4 b)}^{0} + 8 \cdot {(3 a)}^{7} \cdot {(4 b)}^{1} + \dots + 1 \cdot {(3 a)}^{0} \cdot {(4 b)}^{8}$

Step 4

Simplify the polynomial result.

$6561 a^{8} + 69984 a^{7} b + 326592 a^{6} b^{2} + 870912 a^{5} b^{3} + 1451520 a^{4} b^{4} + 1548288 a^{3} b^{5} + 1032192 a^{2} b^{6} + 393216 a b^{7} + 65536 b^{8}$

Expand Using the Binomial Theorem (3a+4b)^8

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