Question

7.6
score: 7/19 answered: 7/19
question 8
use the binomial formula to expand the following:
(2x - y)^6 =
question help: video ebook

Explanation:

Step1: Recall Binomial Theorem

The binomial theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\), \(a = 2x\), \(b=-y\), and \(n = 6\).

Step2: Calculate each term for \(k = 0\) to \(k = 6\)

• For \(k = 0\):

\(\binom{6}{0}(2x)^{6}(-y)^{0}=\frac{6!}{0!6!}(64x^{6})(1)=64x^{6}\)

• For \(k = 1\):

\(\binom{6}{1}(2x)^{5}(-y)^{1}=\frac{6!}{1!5!}(32x^{5})(-y)=6\times32x^{5}(-y)=- 192x^{5}y\)

• For \(k = 2\):

\(\binom{6}{2}(2x)^{4}(-y)^{2}=\frac{6!}{2!4!}(16x^{4})(y^{2}) = 15\times16x^{4}y^{2}=240x^{4}y^{2}\)

• For \(k = 3\):

\(\binom{6}{3}(2x)^{3}(-y)^{3}=\frac{6!}{3!3!}(8x^{3})(-y^{3})=20\times8x^{3}(-y^{3})=-160x^{3}y^{3}\)

• For \(k = 4\):

\(\binom{6}{4}(2x)^{2}(-y)^{4}=\frac{6!}{4!2!}(4x^{2})(y^{4})=15\times4x^{2}y^{4}=60x^{2}y^{4}\)

• For \(k = 5\):

\(\binom{6}{5}(2x)^{1}(-y)^{5}=\frac{6!}{5!1!}(2x)(-y^{5})=6\times2x(-y^{5})=-12xy^{5}\)

• For \(k = 6\):

\(\binom{6}{6}(2x)^{0}(-y)^{6}=\frac{6!}{6!0!}(1)(y^{6})=1\times1\times y^{6}=y^{6}\)

Step3: Sum all terms

\((2x - y)^{6}=64x^{6}-192x^{5}y + 240x^{4}y^{2}-160x^{3}y^{3}+60x^{2}y^{4}-12xy^{5}+y^{6}\)

Answer:

\(64x^{6}-192x^{5}y + 240x^{4}y^{2}-160x^{3}y^{3}+60x^{2}y^{4}-12xy^{5}+y^{6}\)