Solve sqrt[3]{24}-5sqrt[6]{27}+3sqrt{675}-sqrt{48}=

$$\sqrt[ 3 ] { 24 } - 5 \sqrt[ 6 ] { 27 } + 3 \sqrt { 675 } - \sqrt { 48 } =$$

Evaluate

$$2\sqrt[3]{3}+36\sqrt{3}\approx 65.238328213$$

Solution Steps

Rewrite \sqrt[6]{27} as \sqrt[6]{3^{3}}. Convert from radical to exponential form and cancel out 3 in the exponent. Convert back to radical form.

$$\sqrt[6]{27}=\sqrt[6]{3^{3}}=3^{\frac{3}{6}}=3^{\frac{1}{2}}=\sqrt{3}$$

Insert the obtained value back in the expression.

$$\sqrt[3]{24}-5\sqrt{3}+3\sqrt{675}-\sqrt{48}$$

Factor 675=15^{2}\times 3. Rewrite the square root of the product \sqrt{15^{2}\times 3} as the product of square roots \sqrt{15^{2}}\sqrt{3}. Take the square root of 15^{2}.

$$\sqrt[3]{24}-5\sqrt{3}+3\times 15\sqrt{3}-\sqrt{48}$$

Multiply 3 and 15 to get 45.

$$\sqrt[3]{24}-5\sqrt{3}+45\sqrt{3}-\sqrt{48}$$

Combine -5\sqrt{3} and 45\sqrt{3} to get 40\sqrt{3}.

$$\sqrt[3]{24}+40\sqrt{3}-\sqrt{48}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\sqrt[3]{24}+40\sqrt{3}-4\sqrt{3}$$

Combine 40\sqrt{3} and -4\sqrt{3} to get 36\sqrt{3}.

$$\sqrt[3]{24}+36\sqrt{3}$$