4.b) giải như sau:
[tex]M=a^{3}+b^{3}= \sqrt{8-4\sqrt{3}}-\frac{4}{\sqrt{2}+\sqrt{6}}[/tex]
Phân tích [tex]8-4\sqrt{3}=\left (\sqrt{2} \right )^{2}-2.2.\sqrt{3}+\left ( \sqrt{6} \right )^{2}=\left (\sqrt{2} \right )^{2}-2.\sqrt{12}+\left ( \sqrt{6} \right )^{2}=\left (\sqrt{2} \right )^{2}-2.\sqrt{2}.\sqrt{6}+\left ( \sqrt{6} \right )^{2}=\left ( \sqrt{2} +\sqrt{6}\right )^{2}[/tex]
Do đó: [tex]M=\frac{\sqrt{\left ( \sqrt{2}+\sqrt{6} \right )^{2}}.\left ( \sqrt{2}+\sqrt{6} \right )-4}{\sqrt{2}+\sqrt{6}}=\frac{\left (\sqrt{2}+\sqrt{6} \right )^{2}+4}{\sqrt{2}+\sqrt{6}}[/tex]
[tex]M=\frac{8+2\sqrt{12}+4}{\sqrt{2}+\sqrt{6}}=\frac{12+2\sqrt{12}}{\sqrt{2}+\sqrt{6}}=\frac{\sqrt{12}\left ( \sqrt{12}+2 \right )}{\sqrt{2}\left ( 1+\sqrt{3} \right )}=\frac{\sqrt{6}\left ( 2\sqrt{3}+2 \right )}{1+\sqrt{3}}=\frac{2.\sqrt{6}.\left (\sqrt{3}+1 \right )}{\sqrt{3}+1}=2\sqrt{6}[/tex]
Vậy [tex]M=2\sqrt{6}[/tex]