Solve 13/2sqrt{96}+sqrt[2]{486}-2sqrt[3]{6}+9sqrt[4]{243}

$$1 \frac { 3 } { 2 } \sqrt { 96 } + \sqrt[ 2 ] { 486 } - 2 \sqrt[ 3 ] { 6 } + 9 \sqrt[ 4 ] { 243 }$$

Evaluate

$$19\sqrt{6}+27\sqrt[4]{3}-2\sqrt[3]{6}\approx 78.440062277$$

Solution Steps

Multiply 1 and 2 to get 2.

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

Add 2 and 3 to get 5.

$$\frac{5}{2}\sqrt{96}+\sqrt[2]{486}-2\sqrt[3]{6}+9\sqrt[4]{243}$$

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}.

$$\frac{5}{2}\times 4\sqrt{6}+\sqrt[2]{486}-2\sqrt[3]{6}+9\sqrt[4]{243}$$

Multiply \frac{5}{2} and 4 to get 10.

$$10\sqrt{6}+\sqrt[2]{486}-2\sqrt[3]{6}+9\sqrt[4]{243}$$

Factor

$$19 \sqrt{6} + 27 \sqrt[4]{3} - 2 \sqrt[3]{6} = 78.440062277$$