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{(1 + x + x^{2})}^{3}

${(1 + x + x^{2})}^{3}$

Langkah 1

Gunakan perluasan trinomial untuk menentukan setiap suku. Teorema trinomial menyatakan

{(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}

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

{}_{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!} \cdot 1^{3 - m} x^{m - k} {(x^{2})}^{k}

$\sum_{m = 0}^{3} \sum_{k = 0}^{m} \frac{3!}{(3 - m)! (m - k)! k!} \cdot 1^{3 - m} x^{m - k} {(x^{2})}^{k}$

Langkah 2

Perluas penjumlahannya.

\frac{3!}{(3 + 0)! (0 + 0)! 0!} \cdot 1^{3 - 0} x^{0 - 0} {(x^{2})}^{0} + \frac{3!}{(3 - 1)! (1 + 0)! 0!} \cdot 1^{3 - 1 \cdot 1} x^{1 - 0} {(x^{2})}^{0} + \dots + \frac{3!}{(3 - 3)! (3 - 3)! 3!} \cdot 1^{3 - 1 \cdot 3} x^{3 - 1 \cdot 3} {(x^{2})}^{3}

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

Langkah 3

Sederhanakan hasilnya.

1 + 3 x + 6 x^{2} + 6 x^{4} + 7 x^{3} + 3 x^{5} + x^{6}

$1 + 3 x + 6 x^{2} + 6 x^{4} + 7 x^{3} + 3 x^{5} + x^{6}$

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