Solve (sqrt{3}+3)^2-2sqrt{27}

$$( \sqrt { 3 } + 3 ) ^ { 2 } - 2 \sqrt { 27 }$$

Evaluate

$$12$$

Solution Steps

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

$$\left(\sqrt{3}\right)^{2}+6\sqrt{3}+9-2\sqrt{27}$$

The square of \sqrt{3} is 3.

$$3+6\sqrt{3}+9-2\sqrt{27}$$

Add 3 and 9 to get 12.

$$12+6\sqrt{3}-2\sqrt{27}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$12+6\sqrt{3}-2\times 3\sqrt{3}$$

Multiply -2 and 3 to get -6.

$$12+6\sqrt{3}-6\sqrt{3}$$

Combine 6\sqrt{3} and -6\sqrt{3} to get 0.

$$12$$

Factor

$$2^{2}\times 3$$