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

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

Evaluate

$$3\left(a^{2}-2b^{2}\right)$$

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}.

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

Expand \left(\sqrt{3}a\right)^{2}.

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{6} is 6.

$$3a^{2}-6b^{2}$$

Differentiate w.r.t. a

$$6a$$