# How do you evaluate 2√44x^3 -√7 - √99x^3 +√63?

Nov 8, 2015

x sqrt(11x)+2sqrt7)

#### Explanation:

$44 = 4 \times 11 \text{ and 4 is } {2}^{2}$
$99 = 9 \times 11 \text{ and 9 is } {3}^{2}$

Both 4 and 9 have square roots

$63 = 3 \times 21 = 3 \times 3 \times 7 = {3}^{2} \times 7$

$2 \sqrt{44 {x}^{3}} - \sqrt{7} - \sqrt{99 {x}^{3}} + \sqrt{63} \text{ }$ becomes

$2 \sqrt{{2}^{2} \times 11 {x}^{3}} - \sqrt{7} - \sqrt{{3}^{2} \times 11 {x}^{3}} + \sqrt{{3}^{2} \times 7}$

$4 \sqrt{11 {x}^{3}} - 3 \sqrt{11 {x}^{3}} + 3 \sqrt{7} - \sqrt{7}$

$\sqrt{11 {x}^{3}} + 2 \sqrt{7}$

But ${x}^{3} = {x}^{2} x$ giving:

$x \sqrt{11 x} + 2 \sqrt{7}$

There are no more squares that can be factored out of 11 and 7 so this is the final answer/