Solve frac{7+sqrt{6}}{7-sqrt{6}}-frac{7-sqrt{6}}{7+sqrt{6}}

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

Evaluate

$$\frac{28\sqrt{6}}{43}\approx 1.595016577$$

Short Solution Steps

Rationalize the denominator of \frac{7+\sqrt{6}}{7-\sqrt{6}} by multiplying numerator and denominator by 7+\sqrt{6}.

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

Consider \left(7-\sqrt{6}\right)\left(7+\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 7. Square \sqrt{6}.

$$\frac{\left(7+\sqrt{6}\right)\left(7+\sqrt{6}\right)}{49-6}-\frac{7-\sqrt{6}}{7+\sqrt{6}}$$

Subtract 6 from 49 to get 43.

$$\frac{\left(7+\sqrt{6}\right)\left(7+\sqrt{6}\right)}{43}-\frac{7-\sqrt{6}}{7+\sqrt{6}}$$

Multiply 7+\sqrt{6} and 7+\sqrt{6} to get \left(7+\sqrt{6}\right)^{2}.

$$\frac{\left(7+\sqrt{6}\right)^{2}}{43}-\frac{7-\sqrt{6}}{7+\sqrt{6}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(7+\sqrt{6}\right)^{2}.

$$\frac{49+14\sqrt{6}+\left(\sqrt{6}\right)^{2}}{43}-\frac{7-\sqrt{6}}{7+\sqrt{6}}$$

The square of \sqrt{6} is 6.

$$\frac{49+14\sqrt{6}+6}{43}-\frac{7-\sqrt{6}}{7+\sqrt{6}}$$

Add 49 and 6 to get 55.

$$\frac{55+14\sqrt{6}}{43}-\frac{7-\sqrt{6}}{7+\sqrt{6}}$$

Rationalize the denominator of \frac{7-\sqrt{6}}{7+\sqrt{6}} by multiplying numerator and denominator by 7-\sqrt{6}.

$$\frac{55+14\sqrt{6}}{43}-\frac{\left(7-\sqrt{6}\right)\left(7-\sqrt{6}\right)}{\left(7+\sqrt{6}\right)\left(7-\sqrt{6}\right)}$$

Consider \left(7+\sqrt{6}\right)\left(7-\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{55+14\sqrt{6}}{43}-\frac{\left(7-\sqrt{6}\right)\left(7-\sqrt{6}\right)}{7^{2}-\left(\sqrt{6}\right)^{2}}$$

Square 7. Square \sqrt{6}.

$$\frac{55+14\sqrt{6}}{43}-\frac{\left(7-\sqrt{6}\right)\left(7-\sqrt{6}\right)}{49-6}$$

Subtract 6 from 49 to get 43.

$$\frac{55+14\sqrt{6}}{43}-\frac{\left(7-\sqrt{6}\right)\left(7-\sqrt{6}\right)}{43}$$

Multiply 7-\sqrt{6} and 7-\sqrt{6} to get \left(7-\sqrt{6}\right)^{2}.

$$\frac{55+14\sqrt{6}}{43}-\frac{\left(7-\sqrt{6}\right)^{2}}{43}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-\sqrt{6}\right)^{2}.

$$\frac{55+14\sqrt{6}}{43}-\frac{49-14\sqrt{6}+\left(\sqrt{6}\right)^{2}}{43}$$

The square of \sqrt{6} is 6.

$$\frac{55+14\sqrt{6}}{43}-\frac{49-14\sqrt{6}+6}{43}$$

Add 49 and 6 to get 55.

$$\frac{55+14\sqrt{6}}{43}-\frac{55-14\sqrt{6}}{43}$$

Since \frac{55+14\sqrt{6}}{43} and \frac{55-14\sqrt{6}}{43} have the same denominator, subtract them by subtracting their numerators.

$$\frac{55+14\sqrt{6}-\left(55-14\sqrt{6}\right)}{43}$$

Do the multiplications in 55+14\sqrt{6}-\left(55-14\sqrt{6}\right).

$$\frac{55+14\sqrt{6}-55+14\sqrt{6}}{43}$$

Do the calculations in 55+14\sqrt{6}-55+14\sqrt{6}.

$$\frac{28\sqrt{6}}{43}$$