Solve (2sqrt{6}-sqrt{3})(2sqrt{6}-sqrt{3})

$$( 2 \sqrt { 6 } - \sqrt { 3 } ) ( 2 \sqrt { 6 } - \sqrt { 3 } )$$

Evaluate

$$27-12\sqrt{2}\approx 10.029437252$$

Solution Steps

Multiply 2\sqrt{6}-\sqrt{3} and 2\sqrt{6}-\sqrt{3} to get \left(2\sqrt{6}-\sqrt{3}\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{6}-\sqrt{3}\right)^{2}.

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

The square of \sqrt{6} is 6.

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

Multiply 4 and 6 to get 24.

$$24-4\sqrt{6}\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

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}.

$$24-4\sqrt{3}\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$24-4\times 3\sqrt{2}+\left(\sqrt{3}\right)^{2}$$

Multiply -4 and 3 to get -12.

$$24-12\sqrt{2}+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$24-12\sqrt{2}+3$$

Add 24 and 3 to get 27.

$$27-12\sqrt{2}$$

Factor

$$3 {(9 - 4 \sqrt{2})} = 10.029437252$$