Solve frac{1+6sqrt{6}}{1-6sqrt{6}}

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

Evaluate

$$\frac{-12\sqrt{6}-217}{215}\approx -1.146018032$$

Solution Steps

Rationalize the denominator of \frac{1+6\sqrt{6}}{1-6\sqrt{6}} by multiplying numerator and denominator by 1+6\sqrt{6}.

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

Consider \left(1-6\sqrt{6}\right)\left(1+6\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}.

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

Multiply 1+6\sqrt{6} and 1+6\sqrt{6} to get \left(1+6\sqrt{6}\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+6\sqrt{6}\right)^{2}.

$$\frac{1+12\sqrt{6}+36\left(\sqrt{6}\right)^{2}}{1^{2}-\left(-6\sqrt{6}\right)^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{1+12\sqrt{6}+36\times 6}{1^{2}-\left(-6\sqrt{6}\right)^{2}}$$

Multiply 36 and 6 to get 216.

$$\frac{1+12\sqrt{6}+216}{1^{2}-\left(-6\sqrt{6}\right)^{2}}$$

Add 1 and 216 to get 217.

$$\frac{217+12\sqrt{6}}{1^{2}-\left(-6\sqrt{6}\right)^{2}}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{217+12\sqrt{6}}{1-\left(-6\sqrt{6}\right)^{2}}$$

Expand \left(-6\sqrt{6}\right)^{2}.

$$\frac{217+12\sqrt{6}}{1-\left(-6\right)^{2}\left(\sqrt{6}\right)^{2}}$$

Calculate -6 to the power of 2 and get 36.

$$\frac{217+12\sqrt{6}}{1-36\left(\sqrt{6}\right)^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{217+12\sqrt{6}}{1-36\times 6}$$

Multiply 36 and 6 to get 216.

$$\frac{217+12\sqrt{6}}{1-216}$$

Subtract 216 from 1 to get -215.

$$\frac{217+12\sqrt{6}}{-215}$$

Multiply both numerator and denominator by -1.

$$\frac{-217-12\sqrt{6}}{215}$$