# How do I simplify (3sqrt3 - 1)^2 / (2sqrt3 - 3) ?

##### 3 Answers
Apr 14, 2018

$\frac{2 \left(19 \sqrt{3} + 24\right)}{3}$

#### Explanation:

${\left(3 \sqrt{3} - 1\right)}^{2} / \left(2 \sqrt{3} - 3\right)$

$= {\left(3 \sqrt{3} - 1\right)}^{2} / \left(2 \sqrt{3} - 3\right) \cdot \frac{2 \sqrt{3} + 3}{2 \sqrt{3} + 3}$

$= \frac{\left(27 - 6 \sqrt{3} + 1\right) \left(2 \sqrt{3} + 3\right)}{{\left(2 \sqrt{3}\right)}^{2} - {\left(3\right)}^{2}}$

$= \frac{2 \left(14 - 3 \sqrt{3}\right) \left(2 \sqrt{3} + 3\right)}{12 - 9}$

$= \frac{2 \left(28 \sqrt{3} + 42 - 18 - 9 \sqrt{3}\right)}{3}$

$= \frac{2 \left(19 \sqrt{3} + 24\right)}{3}$

Apr 14, 2018

$16 + \frac{38}{3} \sqrt{3}$

#### Explanation:

$\text{the first thing to do is expand the numerator using FOIL}$

$\text{note that } \sqrt{a} \times \sqrt{a} = a$

$\Rightarrow {\left(3 \sqrt{3} - 1\right)}^{2} = \left(3 \sqrt{3} - 1\right) \left(3 \sqrt{3} - 1\right)$

$= 27 - 3 \sqrt{3} - 3 \sqrt{3} + 1$

$= 28 - 6 \sqrt{3}$

$\Rightarrow \frac{{\left(3 \sqrt{3} - 1\right)}^{2}}{2 \sqrt{3} - 3} = \frac{28 - 6 \sqrt{3}}{2 \sqrt{3} - 3}$

$\text{the next step is to "color(blue)"rationalise the denominator}$
$\text{that is, eliminate the radical}$

$\text{to do this multiply numerator/denominator by the}$
$\textcolor{b l u e}{\text{conjugate ""of the denominator}}$

$\text{the conjugate of "2sqrt3-3" is } 2 \sqrt{3} \textcolor{red}{+} 3$

rArr((28-6sqrt3)(2sqrt3+3))/((2sqrt3-3)(2sqrt3+3)

$\text{expand numerator/denominator using FOIL}$

$= \frac{56 \sqrt{3} + 84 - 36 - 18 \sqrt{3}}{12 \cancel{+ 6 \sqrt{3}} \cancel{- 6 \sqrt{3}} - 9}$

$= \frac{48 + 38 \sqrt{3}}{3}$

$= 16 + \frac{38}{3} \sqrt{3}$

Apr 14, 2018

Given expression

$\frac{{\left(3 \sqrt{3} - 1\right)}^{2}}{2 \sqrt{3} - 3}$

Expanding the numerator using the identity

${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$, we get

$\frac{{\left(3 \sqrt{3}\right)}^{2} - 2 \times 3 \sqrt{3} \times 1 + {1}^{2}}{2 \sqrt{3} - 3}$
$\implies \frac{27 - 6 \sqrt{3} + 1}{2 \sqrt{3} - 3}$
$\implies \frac{28 - 6 \sqrt{3}}{2 \sqrt{3} - 3}$

Rationalizing the denominator and using the identity $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$, we get

$\frac{28 - 6 \sqrt{3}}{2 \sqrt{3} - 3} \times \frac{2 \sqrt{3} + 3}{2 \sqrt{3} + 3}$
$\implies \frac{\left(28 - 6 \sqrt{3}\right) \left(2 \sqrt{3} + 3\right)}{{\left(2 \sqrt{3}\right)}^{2} - {\left(3\right)}^{2}}$
$\implies \frac{\left(56 \sqrt{3} + 84 - 36 - 18 \sqrt{3}\right)}{{\left(2 \sqrt{3}\right)}^{2} - {\left(3\right)}^{2}}$
=>((38sqrt3+48))/((12-9)
$\implies \frac{2}{3} \left(19 \sqrt{3} + 24\right)$