Solve 3/2sqrt{96}+sqrt[2]{486}-2sqrt[3]{6}+9sqrt[4]{243}

$$\frac{ 3 }{ 2 } \sqrt{ 96 } + \sqrt[ 2 ]{ 486 } -2 \sqrt[ 3 ]{ 6 } +9 \sqrt[ 4 ]{ 243 }$$

Evaluate

$$15\sqrt{6}+27\sqrt[4]{3}-2\sqrt[3]{6}\approx 68.642103306$$

Solution Steps

Factor 96=4^{2}\times 6. Rewrite the square root of the product \sqrt{4^{2}\times 6} as the product of square roots \sqrt{4^{2}}\sqrt{6}. Take the square root of 4^{2}.

$$\frac{3}{2}\times 4\sqrt{6}+\sqrt[2]{486}-2\sqrt[3]{6}+9\sqrt[4]{243}$$

Multiply \frac{3}{2} and 4 to get 6.

$$6\sqrt{6}+\sqrt[2]{486}-2\sqrt[3]{6}+9\sqrt[4]{243}$$

Factor

$$15 \sqrt{6} + 27 \sqrt[4]{3} - 2 \sqrt[3]{6} = 68.642103306$$