a) Ta có: \(\sqrt{\left(6+2\sqrt{5}\right)^3}-\sqrt{\left(6-2\sqrt{5}\right)^3}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}^3-\sqrt{\left(\sqrt{5}-1\right)^2}^3\)
\(=\left|\left(\sqrt{5}+1\right)^3\right|-\left|\left(\sqrt{5}-1\right)^3\right|\)
\(=\left(\sqrt{5}+1\right)^3-\left(\sqrt{5}-1\right)^3\)
\(=\left[\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\right]\cdot\left[\left(\sqrt{5}+1\right)^2+\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)+\left(\sqrt{5}-1\right)^2\right]\)
\(=\left(\sqrt{5}+1-\sqrt{5}+1\right)\cdot\left(6+2\sqrt{5}+5-1+6-2\sqrt{5}\right)\)
\(=2\cdot16=32\)
b) Ta có: \(\frac{a^3-2\sqrt{2}}{a-\sqrt{2}}\)
\(=\frac{\left(a-\sqrt{2}\right)\left(a^2+a\sqrt{2}+2\right)}{a-\sqrt{2}}=a^2+\sqrt{2}\cdot a+2\)
