# How do you use the binomial theorem to expand and simplify the expression (1/x+2y)^6?

Feb 11, 2018

${x}^{-} 6 + 12 y {x}^{-} 5 + 60 {y}^{2} {x}^{-} 4 + 160 {y}^{3} {x}^{-} 3 + 240 {y}^{4} {x}^{-} 2 + 192 {y}^{5} {x}^{-} 1 + 64 {y}^{6}$

#### Explanation:

$1$
$1$ $1$
$1$ $2$ $1$
$1$ $3$3 $1$
$1$ $4$ $6$ $4$ $1$
$1$ $5$ $10$ $10$ $5$ $1$
$1$ $6$ $15$ $20$ $15$ $6$ $1$ < 6th power

${\left(\frac{1}{x} + 2 y\right)}^{6} = 1 \left({\left(\frac{1}{x}\right)}^{6}\right) + 6 \left({\left(\frac{1}{x}\right)}^{5} 2 y\right) + 15 \left({\left(\frac{1}{x}\right)}^{4} {\left(2 y\right)}^{2}\right) + 20 \left({\left(\frac{1}{x}\right)}^{3} {\left(2 y\right)}^{3}\right) + 15 \left({\left(\frac{1}{x}\right)}^{2} {\left(2 y\right)}^{4}\right) + 6 \left(\frac{1}{x} \left(2 {y}^{5}\right)\right) + 1 \left({\left(2 y\right)}^{6}\right)$

${\left(\frac{1}{x}\right)}^{6} = \frac{1}{x} ^ 6 = {x}^{-} 6$

$6 \left({\left(\frac{1}{x}\right)}^{5} 2 y\right) = \frac{1}{x} ^ 5 \cdot 12 y = 12 y {x}^{-} 5$

$15 \left({\left(\frac{1}{x}\right)}^{4} {\left(2 y\right)}^{2}\right) = \frac{1}{x} ^ 4 \cdot 60 {y}^{2} = 60 {y}^{2} {x}^{-} 4$

$20 \left({\left(\frac{1}{x}\right)}^{3} {\left(2 y\right)}^{3}\right) = \frac{1}{x} ^ 3 \cdot 160 {y}^{3} = 160 {y}^{3} {x}^{-} 3$

$15 \left({\left(\frac{1}{x}\right)}^{2} {\left(2 y\right)}^{4}\right) = \frac{1}{x} ^ 2 \cdot 240 {y}^{4} = 240 {y}^{4} {x}^{-} 2$

$6 \left(\frac{1}{x} {\left(2 y\right)}^{5}\right) = \frac{1}{x} \cdot 192 {y}^{5} = 192 {y}^{5} {x}^{-} 1$

${\left(2 y\right)}^{6} = 64 {y}^{6}$

total:

${x}^{-} 6 + 12 y {x}^{-} 5 + 60 {y}^{2} {x}^{-} 4 + 160 {y}^{3} {x}^{-} 3 + 240 {y}^{4} {x}^{-} 2 + 192 {y}^{5} {x}^{-} 1 + 64 {y}^{6}$

there are no common factors between all of the terms.