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4\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{6}+\left(\sqrt{6}\right)^{2}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2\sqrt{3}+\sqrt{6}\right)^{2}.
4\times 3+4\sqrt{3}\sqrt{6}+\left(\sqrt{6}\right)^{2}
The square of \sqrt{3} is 3.
12+4\sqrt{3}\sqrt{6}+\left(\sqrt{6}\right)^{2}
Multiply 4 and 3 to get 12.
12+4\sqrt{3}\sqrt{3}\sqrt{2}+\left(\sqrt{6}\right)^{2}
Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.
12+4\times 3\sqrt{2}+\left(\sqrt{6}\right)^{2}
Multiply \sqrt{3} and \sqrt{3} to get 3.
12+12\sqrt{2}+\left(\sqrt{6}\right)^{2}
Multiply 4 and 3 to get 12.
12+12\sqrt{2}+6
The square of \sqrt{6} is 6.
18+12\sqrt{2}
Add 12 and 6 to get 18.
4\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{6}+\left(\sqrt{6}\right)^{2}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2\sqrt{3}+\sqrt{6}\right)^{2}.
4\times 3+4\sqrt{3}\sqrt{6}+\left(\sqrt{6}\right)^{2}
The square of \sqrt{3} is 3.
12+4\sqrt{3}\sqrt{6}+\left(\sqrt{6}\right)^{2}
Multiply 4 and 3 to get 12.
12+4\sqrt{3}\sqrt{3}\sqrt{2}+\left(\sqrt{6}\right)^{2}
Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.
12+4\times 3\sqrt{2}+\left(\sqrt{6}\right)^{2}
Multiply \sqrt{3} and \sqrt{3} to get 3.
12+12\sqrt{2}+\left(\sqrt{6}\right)^{2}
Multiply 4 and 3 to get 12.
12+12\sqrt{2}+6
The square of \sqrt{6} is 6.
18+12\sqrt{2}
Add 12 and 6 to get 18.