Expand Using the Binomial Theorem (2y^2-1)^6

${(2 y^{2} - 1)}^{6}$

Step 1

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

$\sum_{k = 0}^{6} \frac{6!}{(6 - k)! k!} \cdot {(2 y^{2})}^{6 - k} \cdot {(- 1)}^{k}$

Step 2

Expand the summation.

$\frac{6!}{(6 - 0)! 0!} \cdot {(2 y^{2})}^{6 - 0} \cdot {(- 1)}^{0} + \frac{6!}{(6 - 1)! 1!} \cdot {(2 y^{2})}^{6 - 1} \cdot {(- 1)}^{1} + \dots + \frac{6!}{(6 - 6)! 6!} \cdot {(2 y^{2})}^{6 - 6} \cdot {(- 1)}^{6}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 y^{2})}^{6} \cdot {(- 1)}^{0} + 6 \cdot {(2 y^{2})}^{5} \cdot {(- 1)}^{1} + \dots + 1 \cdot {(2 y^{2})}^{0} \cdot {(- 1)}^{6}$

Step 4

Simplify the polynomial result.

$64 y^{12} - 192 y^{10} + 240 y^{8} - 160 y^{6} + 60 y^{4} - 12 y^{2} + 1$

Expand Using the Binomial Theorem (2y^2-1)^6

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