Solve (sqrt{6}-sqrt{3})^2*(9+6sqrt{2})

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

Evaluate

$$9$$

Solution Steps

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

$$\left(\left(\sqrt{6}\right)^{2}-2\sqrt{6}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)\left(9+6\sqrt{2}\right)$$

The square of \sqrt{6} is 6.

$$\left(6-2\sqrt{6}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)\left(9+6\sqrt{2}\right)$$

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}.

$$\left(6-2\sqrt{3}\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)\left(9+6\sqrt{2}\right)$$

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

$$\left(6-2\times 3\sqrt{2}+\left(\sqrt{3}\right)^{2}\right)\left(9+6\sqrt{2}\right)$$

Multiply -2 and 3 to get -6.

$$\left(6-6\sqrt{2}+\left(\sqrt{3}\right)^{2}\right)\left(9+6\sqrt{2}\right)$$

The square of \sqrt{3} is 3.

$$\left(6-6\sqrt{2}+3\right)\left(9+6\sqrt{2}\right)$$

Add 6 and 3 to get 9.

$$\left(9-6\sqrt{2}\right)\left(9+6\sqrt{2}\right)$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 9.

$$81-\left(6\sqrt{2}\right)^{2}$$

Expand \left(6\sqrt{2}\right)^{2}.

$$81-6^{2}\left(\sqrt{2}\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$81-36\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$81-36\times 2$$

Multiply 36 and 2 to get 72.

$$81-72$$

Subtract 72 from 81 to get 9.

$$9$$

Factor

$$3^{2}$$