Examples

{(a + b + c)}^{3}

${(a + b + c)}^{3}$

Step 1

Use the trinomial expansion theorem to find each term. The trinomial theorem states

{(a + b + c)}^{n} = \sum_{m = 0}^{n} \sum_{k = 0}^{m} {}_{n}C_{m} {}_{m}C_{k} a^{n - m} b^{m - k} c^{k}

${(a + b + c)}^{n} = \sum_{m = 0}^{n} \sum_{k = 0}^{m} {}_{n}C_{m} {}_{m}C_{k} a^{n - m} b^{m - k} c^{k}$ where

{}_{n}C_{m} {}_{m}C_{k} = \frac{n!}{(n - m)! (m - k)! k!}

${}_{n}C_{m} {}_{m}C_{k} = \frac{n!}{(n - m)! (m - k)! k!}$ .

\sum_{m = 0}^{3} \sum_{k = 0}^{m} \frac{3!}{(3 - m)! (m - k)! k!} a^{3 - m} b^{m - k} c^{k}

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

Step 2

Expand the summation.

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

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

Step 3

Simplify the result.

a^{3} + 3 a^{2} b + 3 a^{2} c + 3 a b^{2} + 6 a b c + 3 a c^{2} + b^{3} + 3 b^{2} c + 3 b c^{2} + c^{3}

$a^{3} + 3 a^{2} b + 3 a^{2} c + 3 a b^{2} + 6 a b c + 3 a c^{2} + b^{3} + 3 b^{2} c + 3 b c^{2} + c^{3}$

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