How do you use the binomial theorem to expand (2sqrtt-1)^3?

Jan 1, 2018

${\left(2 \sqrt{t} - 1\right)}^{3} = 8 t \sqrt{t} - 12 t + 6 \sqrt{t} - 1$

Explanation:

To find the binomial coefficients, I will use Pascal's triangle:
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a} 1$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a} 1 \textcolor{w h i t e}{a a a} 1$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a} 1 \textcolor{w h i t e}{a a a} 2 \textcolor{w h i t e}{a a a} 1$
$\textcolor{w h i t e}{a a a a a a a a a a a a a} 1 \textcolor{w h i t e}{a a a} 3 \textcolor{w h i t e}{a a a} 3 \textcolor{w h i t e}{a a a} 1$

Looking at the 4th row (mind you, we're counting from $0$), we can see that the coefficients will be $1 , 3 , 3 , 1$.

The degree of the left term decreases from left to right and the degree of the right term increases from left to right. This means our expansion will be:
${\left(2 \sqrt{t} - 1\right)}^{3} = {\left(2 \sqrt{t}\right)}^{3} + 3 {\left(2 \sqrt{t}\right)}^{2} \left(- 1\right) + 3 \left(2 \sqrt{t}\right) {\left(- 1\right)}^{2} + {\left(- 1\right)}^{3} =$

$= 8 t \sqrt{t} - 12 t + 6 \sqrt{t} - 1$