Solve (sqrt{6}+sqrt{2})^2+(sqrt{6}-sqrt{2})^2=1

$$( \sqrt { 6 } + \sqrt { 2 } ) ^ { 2 } + ( \sqrt { 6 } - \sqrt { 2 } ) ^ { 2 } = 1$$

Verify

Solution Steps

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

$$6+2\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}+\left(\sqrt{6}-\sqrt{2}\right)^{2}=1$$

Multiply 2 and 2 to get 4.

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

The square of \sqrt{2} is 2.

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

Add 6 and 2 to get 8.

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

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

$$8+4\sqrt{3}+6-4\sqrt{3}+2=1$$

Add 6 and 2 to get 8.

$$8+4\sqrt{3}+8-4\sqrt{3}=1$$

Add 8 and 8 to get 16.

$$16+4\sqrt{3}-4\sqrt{3}=1$$

Combine 4\sqrt{3} and -4\sqrt{3} to get 0.

$$16=1$$

Compare 16 and 1.

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