Solve {left(-2-frac{-1-sqrt{61}}{3}right)}^2+{left(1div1/-8-2sqrt{61/3}right)}^2

$${ \left(-2- \frac{ -1- \sqrt{ 61 } }{ 3 } \right) }^{ 2 } + { \left(1 \div \frac{ 1 }{ \frac{ -8-2 \sqrt{ 61 } }{ 3 } } \right) }^{ 2 }$$

Evaluate

$$\frac{22\sqrt{61}+394}{9}\approx 62.869499208$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply -2 times \frac{3}{3}.

$$\left(-\frac{2\times 3}{3}-\frac{-1-\sqrt{61}}{3}\right)^{2}+\left(\frac{1}{\frac{1}{\frac{-8-2\sqrt{61}}{3}}}\right)^{2}$$

Since -\frac{2\times 3}{3} and \frac{-1-\sqrt{61}}{3} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{-2\times 3-\left(-1-\sqrt{61}\right)}{3}\right)^{2}+\left(\frac{1}{\frac{1}{\frac{-8-2\sqrt{61}}{3}}}\right)^{2}$$

Do the multiplications in -2\times 3-\left(-1-\sqrt{61}\right).

$$\left(\frac{-6+1+\sqrt{61}}{3}\right)^{2}+\left(\frac{1}{\frac{1}{\frac{-8-2\sqrt{61}}{3}}}\right)^{2}$$

Do the calculations in -6+1+\sqrt{61}.

$$\left(\frac{-5+\sqrt{61}}{3}\right)^{2}+\left(\frac{1}{\frac{1}{\frac{-8-2\sqrt{61}}{3}}}\right)^{2}$$

To raise \frac{-5+\sqrt{61}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-5+\sqrt{61}\right)^{2}}{3^{2}}+\left(\frac{1}{\frac{1}{\frac{-8-2\sqrt{61}}{3}}}\right)^{2}$$

Divide 1 by \frac{1}{\frac{-8-2\sqrt{61}}{3}} by multiplying 1 by the reciprocal of \frac{1}{\frac{-8-2\sqrt{61}}{3}}.

$$\frac{\left(-5+\sqrt{61}\right)^{2}}{3^{2}}+\left(\frac{-8-2\sqrt{61}}{3}\right)^{2}$$

To raise \frac{-8-2\sqrt{61}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-5+\sqrt{61}\right)^{2}}{3^{2}}+\frac{\left(-8-2\sqrt{61}\right)^{2}}{3^{2}}$$

Since \frac{\left(-5+\sqrt{61}\right)^{2}}{3^{2}} and \frac{\left(-8-2\sqrt{61}\right)^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(-5+\sqrt{61}\right)^{2}+\left(-8-2\sqrt{61}\right)^{2}}{3^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-5+\sqrt{61}\right)^{2}.

$$\frac{25-10\sqrt{61}+\left(\sqrt{61}\right)^{2}}{3^{2}}+\frac{\left(-8-2\sqrt{61}\right)^{2}}{3^{2}}$$

The square of \sqrt{61} is 61.

$$\frac{25-10\sqrt{61}+61}{3^{2}}+\frac{\left(-8-2\sqrt{61}\right)^{2}}{3^{2}}$$

Add 25 and 61 to get 86.

$$\frac{86-10\sqrt{61}}{3^{2}}+\frac{\left(-8-2\sqrt{61}\right)^{2}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{86-10\sqrt{61}}{9}+\frac{\left(-8-2\sqrt{61}\right)^{2}}{3^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-8-2\sqrt{61}\right)^{2}.

$$\frac{86-10\sqrt{61}}{9}+\frac{64+32\sqrt{61}+4\left(\sqrt{61}\right)^{2}}{3^{2}}$$

The square of \sqrt{61} is 61.

$$\frac{86-10\sqrt{61}}{9}+\frac{64+32\sqrt{61}+4\times 61}{3^{2}}$$

Multiply 4 and 61 to get 244.

$$\frac{86-10\sqrt{61}}{9}+\frac{64+32\sqrt{61}+244}{3^{2}}$$

Add 64 and 244 to get 308.

$$\frac{86-10\sqrt{61}}{9}+\frac{308+32\sqrt{61}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{86-10\sqrt{61}}{9}+\frac{308+32\sqrt{61}}{9}$$

Since \frac{86-10\sqrt{61}}{9} and \frac{308+32\sqrt{61}}{9} have the same denominator, add them by adding their numerators.

$$\frac{86-10\sqrt{61}+308+32\sqrt{61}}{9}$$

Do the calculations in 86-10\sqrt{61}+308+32\sqrt{61}.

$$\frac{394+22\sqrt{61}}{9}$$

Factor

$$\frac{2 {(11 \sqrt{61} + 197)}}{9} = 62.869499207771824$$