Expand Using the Binomial Theorem (3x-2y^2)^8

${(3 x - 2 y^{2})}^{8}$

Step 1

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

$\sum_{k = 0}^{8} \frac{8!}{(8 - k)! k!} \cdot {(3 x)}^{8 - k} \cdot {(- 2 y^{2})}^{k}$

Step 2

Expand the summation.

$\frac{8!}{(8 - 0)! 0!} \cdot {(3 x)}^{8 - 0} \cdot {(- 2 y^{2})}^{0} + \frac{8!}{(8 - 1)! 1!} \cdot {(3 x)}^{8 - 1} \cdot {(- 2 y^{2})}^{1} + \dots + \frac{8!}{(8 - 8)! 8!} \cdot {(3 x)}^{8 - 8} \cdot {(- 2 y^{2})}^{8}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(3 x)}^{8} \cdot {(- 2 y^{2})}^{0} + 8 \cdot {(3 x)}^{7} \cdot {(- 2 y^{2})}^{1} + \dots + 1 \cdot {(3 x)}^{0} \cdot {(- 2 y^{2})}^{8}$

Step 4

Simplify the polynomial result.

$6561 x^{8} - 34992 x^{7} y^{2} + 81648 x^{6} y^{4} - 108864 x^{5} y^{6} + 90720 x^{4} y^{8} - 48384 x^{3} y^{10} + 16128 x^{2} y^{12} - 3072 x y^{14} + 256 y^{16}$

Expand Using the Binomial Theorem (3x-2y^2)^8

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