Solve sqrt{9/2}+sqrt{25/8}+sqrt[3]{3000}-8sqrt[3]{3}-sqrt[3]{24}+sqrt{1/8}

$$\sqrt { \frac { 9 } { 2 } } + \sqrt { \frac { 25 } { 8 } } + \sqrt[ 3 ] { 3000 } - 8 \sqrt[ 3 ] { 3 } - \sqrt[ 3 ] { 24 } + \sqrt { \frac { 1 } { 8 } }$$

Evaluate

$$3\sqrt{2}\approx 4.242640687$$

Solution Steps

Rewrite the square root of the division \sqrt{\frac{9}{2}} as the division of square roots \frac{\sqrt{9}}{\sqrt{2}}.

$$\frac{\sqrt{9}}{\sqrt{2}}+\sqrt{\frac{25}{8}}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\sqrt{\frac{1}{8}}$$

Calculate the square root of 9 and get 3.

$$\frac{3}{\sqrt{2}}+\sqrt{\frac{25}{8}}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\sqrt{\frac{1}{8}}$$

Rationalize the denominator of \frac{3}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{3\sqrt{2}}{\left(\sqrt{2}\right)^{2}}+\sqrt{\frac{25}{8}}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\sqrt{\frac{1}{8}}$$

The square of \sqrt{2} is 2.

$$\frac{3\sqrt{2}}{2}+\sqrt{\frac{25}{8}}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\sqrt{\frac{1}{8}}$$

Rewrite the square root of the division \sqrt{\frac{25}{8}} as the division of square roots \frac{\sqrt{25}}{\sqrt{8}}.

$$\frac{3\sqrt{2}}{2}+\frac{\sqrt{25}}{\sqrt{8}}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\sqrt{\frac{1}{8}}$$

Calculate the square root of 25 and get 5.

$$\frac{3\sqrt{2}}{2}+\frac{5}{\sqrt{8}}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\sqrt{\frac{1}{8}}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\frac{3\sqrt{2}}{2}+\frac{5}{2\sqrt{2}}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\sqrt{\frac{1}{8}}$$

Rationalize the denominator of \frac{5}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{3\sqrt{2}}{2}+\frac{5\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\sqrt{\frac{1}{8}}$$

The square of \sqrt{2} is 2.

$$\frac{3\sqrt{2}}{2}+\frac{5\sqrt{2}}{2\times 2}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\sqrt{\frac{1}{8}}$$

Multiply 2 and 2 to get 4.

$$\frac{3\sqrt{2}}{2}+\frac{5\sqrt{2}}{4}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\sqrt{\frac{1}{8}}$$

Combine \frac{3\sqrt{2}}{2} and \frac{5\sqrt{2}}{4} to get \frac{11}{4}\sqrt{2}.

$$\frac{11}{4}\sqrt{2}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\sqrt{\frac{1}{8}}$$

Rewrite the square root of the division \sqrt{\frac{1}{8}} as the division of square roots \frac{\sqrt{1}}{\sqrt{8}}.

$$\frac{11}{4}\sqrt{2}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\frac{\sqrt{1}}{\sqrt{8}}$$

Calculate the square root of 1 and get 1.

$$\frac{11}{4}\sqrt{2}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\frac{1}{\sqrt{8}}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\frac{11}{4}\sqrt{2}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\frac{1}{2\sqrt{2}}$$

Rationalize the denominator of \frac{1}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{11}{4}\sqrt{2}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\frac{\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{11}{4}\sqrt{2}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\frac{\sqrt{2}}{2\times 2}$$

Multiply 2 and 2 to get 4.

$$\frac{11}{4}\sqrt{2}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}+\frac{\sqrt{2}}{4}$$

Combine \frac{11}{4}\sqrt{2} and \frac{\sqrt{2}}{4} to get 3\sqrt{2}.

$$3\sqrt{2}+\sqrt[3]{3000}-8\sqrt[3]{3}-\sqrt[3]{24}$$