# How do you expand the binomial (2x-y)^6 using the binomial theorem?

Aug 29, 2016

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

#### Explanation:

${\left(2 x - y\right)}^{6}$

The binomial theorem states that for any binomial ${\left(a + b\right)}^{n}$, the general expansion is given by ${\left(a + b\right)}^{n} = {\textcolor{w h i t e}{t w o}}_{n} {C}_{r} \times {a}^{n - r} \times {b}^{r}$, where $r$ is in ascending powers from $0$ to $n$ and $n$ is in descending powers from $n$ to $0$.

$= {\textcolor{w h i t e}{t w o}}_{6} {C}_{0} {\left(2 x\right)}^{6} {\left(- y\right)}^{0} + {\textcolor{w h i t e}{t w o}}_{6} {C}_{1} {\left(2 x\right)}^{5} {\left(- y\right)}^{1} + {\textcolor{w h i t e}{t w o}}_{6} {C}_{2} {\left(2 x\right)}^{4} {\left(- y\right)}^{2} + {\textcolor{w h i t e}{t w o}}_{6} {C}_{3} {\left(2 x\right)}^{3} {\left(- y\right)}^{3} + {\textcolor{w h i t e}{t w o}}_{6} {C}_{4} {\left(2 x\right)}^{2} {\left(- y\right)}^{4} + {\textcolor{w h i t e}{t w o}}_{6} {C}_{5} {\left(2 x\right)}^{1} {\left(- y\right)}^{5} + {\textcolor{w h i t e}{t w o}}_{6} {C}_{6} {\left(2 x\right)}^{0} {\left(- y\right)}^{6}$

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

Hopefully this helps!