Solve frac{8sqrt{2}-4sqrt{6}}{2sqrt{2}}-{left(sqrt{3}-1right)}^2

$$\frac{ 8 \sqrt{ 2 } -4 \sqrt{ 6 } }{ 2 \sqrt{ 2 } } - { \left( \sqrt{ 3 } -1 \right) }^{ 2 }$$

Evaluate

$$0$$

Short Solution Steps

Rationalize the denominator of \frac{8\sqrt{2}-4\sqrt{6}}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{\left(8\sqrt{2}-4\sqrt{6}\right)\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}-\left(\sqrt{3}-1\right)^{2}$$

The square of \sqrt{2} is 2.

$$\frac{\left(8\sqrt{2}-4\sqrt{6}\right)\sqrt{2}}{2\times 2}-\left(\sqrt{3}-1\right)^{2}$$

Multiply 2 and 2 to get 4.

$$\frac{\left(8\sqrt{2}-4\sqrt{6}\right)\sqrt{2}}{4}-\left(\sqrt{3}-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

$$\frac{\left(8\sqrt{2}-4\sqrt{6}\right)\sqrt{2}}{4}-\left(\left(\sqrt{3}\right)^{2}-2\sqrt{3}+1\right)$$

The square of \sqrt{3} is 3.

$$\frac{\left(8\sqrt{2}-4\sqrt{6}\right)\sqrt{2}}{4}-\left(3-2\sqrt{3}+1\right)$$

Add 3 and 1 to get 4.

$$\frac{\left(8\sqrt{2}-4\sqrt{6}\right)\sqrt{2}}{4}-\left(4-2\sqrt{3}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4-2\sqrt{3} times \frac{4}{4}.

$$\frac{\left(8\sqrt{2}-4\sqrt{6}\right)\sqrt{2}}{4}-\frac{4\left(4-2\sqrt{3}\right)}{4}$$

Since \frac{\left(8\sqrt{2}-4\sqrt{6}\right)\sqrt{2}}{4} and \frac{4\left(4-2\sqrt{3}\right)}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(8\sqrt{2}-4\sqrt{6}\right)\sqrt{2}-4\left(4-2\sqrt{3}\right)}{4}$$

Do the multiplications in \left(8\sqrt{2}-4\sqrt{6}\right)\sqrt{2}-4\left(4-2\sqrt{3}\right).

$$\frac{16-8\sqrt{3}-16+8\sqrt{3}}{4}$$

Do the calculations in 16-8\sqrt{3}-16+8\sqrt{3}.

$$\frac{0}{4}$$

Zero divided by any non-zero number gives zero.

$$0$$

Factor

$$0$$