[tex]\bf 3\sqrt{128}-2\sqrt{6}\times 4\sqrt{3}\implies 3\sqrt{128}-(2\times 4)\sqrt{18}~~ \begin{cases} 128=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\\ \qquad 2^6\cdot 2\\ \qquad 2^{3\cdot 2}\cdot 2\\ \qquad (2^3)^2\cdot 2\\ \qquad 8^2\cdot 2\\ 18=2\cdot 3\cdot 3\\ \qquad 2\cdot 3^2 \end{cases} \\\\\\ 3\sqrt{8^2\cdot 2}-8\sqrt{2\cdot 3^2}\implies 24\sqrt{2}-24\sqrt{2}\implies \boxed{0}[/tex]
Answer: 0
Step-by-step explanation:
[tex]3\sqrt{128} - (2\sqrt{6})(4\sqrt{3})[/tex]
= [tex]3\sqrt{128} - 2*4\sqrt{6*3}[/tex]
= [tex]3\sqrt{4*4*2*2*2} - 2*4\sqrt{2*3*3}[/tex]
= [tex]3*4*2\sqrt{2} - 2*4*3\sqrt{2}[/tex]
= [tex]24\sqrt{2} - 24\sqrt{2}[/tex]
= 0
