# How do you combine the radical \sqrt{6}-\sqrt{27}+2\sqrt{54}+3\sqrt{48}?

Nov 5, 2014

$\sqrt{2} - \sqrt{27} + 2 \sqrt{54} + 3 \sqrt{48}$

by $27 = {3}^{3}$, $54 = 2 \cdot {3}^{3}$, and $48 = {2}^{4} \cdot 3$,

$= \sqrt{6} - \sqrt{{3}^{3}} + 2 \sqrt{2 \cdot {3}^{3}} + 3 \sqrt{{2}^{4} \cdot 3}$

by pulling out factors out of the square-roots,

$= \sqrt{6} - 3 \sqrt{3} + 6 \sqrt{6} + 12 \sqrt{3}$

by combining the like terms,

$= 7 \sqrt{6} + 9 \sqrt{3}$

I hope that this was helpful.