Solve frac{sqrt{7}-1}{sqrt{7}+1}-frac{sqrt{7}+1}{sqrt{7}-1}

$$\frac { \sqrt { 7 } - 1 } { \sqrt { 7 } + 1 } - \frac { \sqrt { 7 } + 1 } { \sqrt { 7 } - 1 }$$

Evaluate

$$-\frac{2\sqrt{7}}{3}\approx -1.763834207$$

Short Solution Steps

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

$$\frac{\left(\sqrt{7}-1\right)\left(\sqrt{7}-1\right)}{\left(\sqrt{7}+1\right)\left(\sqrt{7}-1\right)}-\frac{\sqrt{7}+1}{\sqrt{7}-1}$$

Consider \left(\sqrt{7}+1\right)\left(\sqrt{7}-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}.

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

Square \sqrt{7}. Square 1.

$$\frac{\left(\sqrt{7}-1\right)\left(\sqrt{7}-1\right)}{7-1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}$$

Subtract 1 from 7 to get 6.

$$\frac{\left(\sqrt{7}-1\right)\left(\sqrt{7}-1\right)}{6}-\frac{\sqrt{7}+1}{\sqrt{7}-1}$$

Multiply \sqrt{7}-1 and \sqrt{7}-1 to get \left(\sqrt{7}-1\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{7}-1\right)^{2}.

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

The square of \sqrt{7} is 7.

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

Add 7 and 1 to get 8.

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

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

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

Consider \left(\sqrt{7}-1\right)\left(\sqrt{7}+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}.

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

Square \sqrt{7}. Square 1.

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

Subtract 1 from 7 to get 6.

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

Multiply \sqrt{7}+1 and \sqrt{7}+1 to get \left(\sqrt{7}+1\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{7}+1\right)^{2}.

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

The square of \sqrt{7} is 7.

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

Add 7 and 1 to get 8.

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

Since \frac{8-2\sqrt{7}}{6} and \frac{8+2\sqrt{7}}{6} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 8-2\sqrt{7}-\left(8+2\sqrt{7}\right).

$$\frac{8-2\sqrt{7}-8-2\sqrt{7}}{6}$$

Do the calculations in 8-2\sqrt{7}-8-2\sqrt{7}.

$$\frac{-4\sqrt{7}}{6}$$

Divide -4\sqrt{7} by 6 to get -\frac{2}{3}\sqrt{7}.

$$-\frac{2}{3}\sqrt{7}$$