Solve (sqrt{3}a)^2+3(sqrt{3}-a)^2

$$( \sqrt { 3 } a ) ^ { 2 } + 3 ( \sqrt { 3 } - a ) ^ { 2 }$$

Evaluate

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

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(\sqrt{3}-a\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Use the distributive property to multiply 3 by 3-2\sqrt{3}a+a^{2}.

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

Combine 3a^{2} and 3a^{2} to get 6a^{2}.

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

Expand

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

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(\sqrt{3}-a\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Use the distributive property to multiply 3 by 3-2\sqrt{3}a+a^{2}.

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

Combine 3a^{2} and 3a^{2} to get 6a^{2}.

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