Solve 1/sqrt{2+1}+7/sqrt{8-1}+sqrt[6]{8}

$$\frac { 1 } { \sqrt { 2 } + 1 } + \frac { 7 } { \sqrt { 8 } - 1 } + \sqrt[ 6 ] { 8 }$$

Evaluate

$$4\sqrt{2}\approx 5.656854249$$

Solution Steps

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

$$\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{7}{\sqrt{8}-1}+\sqrt[6]{8}$$

Consider \left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\sqrt{2}-1}{\left(\sqrt{2}\right)^{2}-1^{2}}+\frac{7}{\sqrt{8}-1}+\sqrt[6]{8}$$

Square \sqrt{2}. Square 1.

$$\frac{\sqrt{2}-1}{2-1}+\frac{7}{\sqrt{8}-1}+\sqrt[6]{8}$$

Subtract 1 from 2 to get 1.

$$\frac{\sqrt{2}-1}{1}+\frac{7}{\sqrt{8}-1}+\sqrt[6]{8}$$

Anything divided by one gives itself.

$$\sqrt{2}-1+\frac{7}{\sqrt{8}-1}+\sqrt[6]{8}$$

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}.

$$\sqrt{2}-1+\frac{7}{2\sqrt{2}-1}+\sqrt[6]{8}$$

Rationalize the denominator of \frac{7}{2\sqrt{2}-1} by multiplying numerator and denominator by 2\sqrt{2}+1.

$$\sqrt{2}-1+\frac{7\left(2\sqrt{2}+1\right)}{\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)}+\sqrt[6]{8}$$

Consider \left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\sqrt{2}-1+\frac{7\left(2\sqrt{2}+1\right)}{\left(2\sqrt{2}\right)^{2}-1^{2}}+\sqrt[6]{8}$$

Expand \left(2\sqrt{2}\right)^{2}.

$$\sqrt{2}-1+\frac{7\left(2\sqrt{2}+1\right)}{2^{2}\left(\sqrt{2}\right)^{2}-1^{2}}+\sqrt[6]{8}$$

Calculate 2 to the power of 2 and get 4.

$$\sqrt{2}-1+\frac{7\left(2\sqrt{2}+1\right)}{4\left(\sqrt{2}\right)^{2}-1^{2}}+\sqrt[6]{8}$$

The square of \sqrt{2} is 2.

$$\sqrt{2}-1+\frac{7\left(2\sqrt{2}+1\right)}{4\times 2-1^{2}}+\sqrt[6]{8}$$

Multiply 4 and 2 to get 8.

$$\sqrt{2}-1+\frac{7\left(2\sqrt{2}+1\right)}{8-1^{2}}+\sqrt[6]{8}$$

Calculate 1 to the power of 2 and get 1.

$$\sqrt{2}-1+\frac{7\left(2\sqrt{2}+1\right)}{8-1}+\sqrt[6]{8}$$

Subtract 1 from 8 to get 7.

$$\sqrt{2}-1+\frac{7\left(2\sqrt{2}+1\right)}{7}+\sqrt[6]{8}$$

Cancel out 7 and 7.

$$\sqrt{2}-1+2\sqrt{2}+1+\sqrt[6]{8}$$

Combine \sqrt{2} and 2\sqrt{2} to get 3\sqrt{2}.

$$3\sqrt{2}-1+1+\sqrt[6]{8}$$

Add -1 and 1 to get 0.

$$3\sqrt{2}+\sqrt[6]{8}$$

Rewrite \sqrt[6]{8} as \sqrt[6]{2^{3}}. Convert from radical to exponential form and cancel out 3 in the exponent. Convert back to radical form.

$$\sqrt[6]{8}=\sqrt[6]{2^{3}}=2^{\frac{3}{6}}=2^{\frac{1}{2}}=\sqrt{2}$$

Insert the obtained value back in the expression.

$$3\sqrt{2}+\sqrt{2}$$

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

$$4\sqrt{2}$$