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\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}+\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}.
6+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{2}\sqrt{3}\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\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}+\left(\sqrt{6}-\sqrt{2}\right)^{2}=1
Multiply \sqrt{2} and \sqrt{2} to get 2.
6+4\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}+2+\left(\sqrt{6}-\sqrt{2}\right)^{2}=1
The square of \sqrt{2} is 2.
8+4\sqrt{3}+\left(\sqrt{6}-\sqrt{2}\right)^{2}=1
Add 6 and 2 to get 8.
8+4\sqrt{3}+\left(\sqrt{6}\right)^{2}-2\sqrt{6}\sqrt{2}+\left(\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}+6-2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}=1
The square of \sqrt{6} is 6.
8+4\sqrt{3}+6-2\sqrt{2}\sqrt{3}\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\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}=1
Multiply \sqrt{2} and \sqrt{2} to get 2.
8+4\sqrt{3}+6-4\sqrt{3}+\left(\sqrt{2}\right)^{2}=1
Multiply -2 and 2 to get -4.
8+4\sqrt{3}+6-4\sqrt{3}+2=1
The square of \sqrt{2} is 2.
8+4\sqrt{3}+8-4\sqrt{3}=1
Add 6 and 2 to get 8.
16+4\sqrt{3}-4\sqrt{3}=1
Add 8 and 8 to get 16.
16=1
Combine 4\sqrt{3} and -4\sqrt{3} to get 0.
\text{false}
Compare 16 and 1.