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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify the polynomial result.

$a^{8} + 8 a^{7} b^{4} + 28 a^{6} b^{8} + 56 a^{5} b^{12} + 70 a^{4} b^{16} + 56 a^{3} b^{20} + 28 a^{2} b^{24} + 8 a b^{28} + b^{32}$

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

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