Answer
$\sqrt{6}+12\sqrt{7}$
Work Step by Step
Simplify each radical to obtain
$3\sqrt{9(6)} -2\sqrt{4(6)}-\sqrt{16(6)}+4\sqrt{9(7)}
\\=3\sqrt{3^2(6)} -2\sqrt{2^2(6)}-\sqrt{4^2(6)}+4\sqrt{3^2(7)}
\\=3\cdot 3\sqrt{6} -2\cdot 2\sqrt{6}-4\sqrt{6}+4\cdot3\sqrt{7}
\\=9\sqrt{6}-4\sqrt{6}-4\sqrt{6}+12\sqrt{7}.$
RECALL:
The distributive property states that for any real numbers a, b, and c,
$(ac-bc)=(a-b)c
\\(ac+bc) = (a+b)c.$
Use the rule above to obtain
$(9-4-4)\sqrt{6}+12\sqrt{7}
\\=\sqrt{6}+12\sqrt{7}.$