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

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

Evaluate

$$6\sqrt{2}\approx 8.485281374$$

Solution Steps

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}, where a=\sqrt{3}+\sqrt{6} and b=3.

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

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

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

The square of \sqrt{3} is 3.

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

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}.

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

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

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

Multiply 2 and 3 to get 6.

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

The square of \sqrt{6} is 6.

$$3+6\sqrt{2}+6-3^{2}$$

Add 3 and 6 to get 9.

$$9+6\sqrt{2}-3^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9+6\sqrt{2}-9$$

Subtract 9 from 9 to get 0.

$$6\sqrt{2}$$