# How do you use the binomial formula to find expand (2x+3)^3?

Feb 5, 2016

${\left(2 x + 3\right)}^{3} = 8 {x}^{3} + 36 {x}^{2} + 54 x + 27$

#### Explanation:

The binomial formula that you need here is

${\left(a + b\right)}^{3} = {a}^{3} + 3 {a}^{2} b + 3 a {b}^{2} + {b}^{3}$

Let me colour the same formula to add clarity:

${\left(\textcolor{red}{a} + \textcolor{b l u e}{b}\right)}^{3} = \textcolor{red}{{a}^{3}} + 3 \textcolor{red}{{a}^{2}} \textcolor{b l u e}{b} + 3 \textcolor{red}{a} \textcolor{b l u e}{{b}^{2}} + \textcolor{b l u e}{{b}^{3}}$

In your case, you would like to expand ${\left(\textcolor{red}{2 x} + \textcolor{b l u e}{3}\right)}^{3}$, so your

$a = 2 x$ and $b = 3$

Now, you need to plug $2 x$ for every occurence of $a$ in the formula and plug $3$ for every occurence of $b$ in the formula:

${\left(\textcolor{red}{2 x} + \textcolor{b l u e}{3}\right)}^{3} = \textcolor{red}{{\left(2 x\right)}^{3}} + 3 \cdot \textcolor{red}{{\left(2 x\right)}^{2}} \cdot \textcolor{b l u e}{3} + 3 \cdot \textcolor{red}{2 x} \cdot \textcolor{b l u e}{{3}^{2}} + \textcolor{b l u e}{{3}^{3}}$

$= {\left(2 x\right)}^{3} + 3 \cdot {\left(2 x\right)}^{2} \cdot 3 + 3 \cdot \left(2 x\right) \cdot {3}^{2} + {3}^{3}$

$= {2}^{3} {x}^{3} + 9 \cdot {2}^{2} {x}^{2} + 3 \cdot 2 x \cdot 9 + 27$

$= 8 {x}^{3} + 36 {x}^{2} + 54 x + 27$