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

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

Evaluate

$$70$$

Solution Steps

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

$$9\left(\sqrt{3}\right)^{2}-12\sqrt{3}+4+\left(6+\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$9\times 3-12\sqrt{3}+4+\left(6+\sqrt{3}\right)^{2}$$

Multiply 9 and 3 to get 27.

$$27-12\sqrt{3}+4+\left(6+\sqrt{3}\right)^{2}$$

Add 27 and 4 to get 31.

$$31-12\sqrt{3}+\left(6+\sqrt{3}\right)^{2}$$

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

$$31-12\sqrt{3}+36+12\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$31-12\sqrt{3}+36+12\sqrt{3}+3$$

Add 36 and 3 to get 39.

$$31-12\sqrt{3}+39+12\sqrt{3}$$

Add 31 and 39 to get 70.

$$70-12\sqrt{3}+12\sqrt{3}$$

Combine -12\sqrt{3} and 12\sqrt{3} to get 0.

$$70$$

Factor

$$2\times 5\times 7$$