Solve {left(sqrt{6}+sqrt{2}right)}^2+2sqrt{2}left(sqrt{6}+sqrt{2}right)

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

Evaluate

$$8\sqrt{3}+12\approx 25.856406461$$

Solution Steps

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

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

The square of \sqrt{6} is 6.

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

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$6+2\sqrt{2}\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}+2\sqrt{2}\left(\sqrt{6}+\sqrt{2}\right)$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$6+2\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}+2\sqrt{2}\left(\sqrt{6}+\sqrt{2}\right)$$

Multiply 2 and 2 to get 4.

$$6+4\sqrt{3}+\left(\sqrt{2}\right)^{2}+2\sqrt{2}\left(\sqrt{6}+\sqrt{2}\right)$$

The square of \sqrt{2} is 2.

$$6+4\sqrt{3}+2+2\sqrt{2}\left(\sqrt{6}+\sqrt{2}\right)$$

Add 6 and 2 to get 8.

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

Use the distributive property to multiply 2\sqrt{2} by \sqrt{6}+\sqrt{2}.

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

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$8+4\sqrt{3}+2\sqrt{2}\sqrt{2}\sqrt{3}+2\left(\sqrt{2}\right)^{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$8+4\sqrt{3}+2\times 2\sqrt{3}+2\left(\sqrt{2}\right)^{2}$$

Multiply 2 and 2 to get 4.

$$8+4\sqrt{3}+4\sqrt{3}+2\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$8+4\sqrt{3}+4\sqrt{3}+2\times 2$$

Multiply 2 and 2 to get 4.

$$8+4\sqrt{3}+4\sqrt{3}+4$$

Combine 4\sqrt{3} and 4\sqrt{3} to get 8\sqrt{3}.

$$8+8\sqrt{3}+4$$

Add 8 and 4 to get 12.

$$12+8\sqrt{3}$$

Factor

$$4 \sqrt{3} {(\sqrt{3} + 2)} = 25.856406461$$