Solve (11-8)^2+(-3-(-3)+3sqrt{3})^2

$$( 11 - 8 ) ^ { 2 } + ( - 3 - ( - 3 ) + 3 \sqrt { 3 } ) ^ { 2 }$$

Evaluate

$$36$$

Solution Steps

Subtract 8 from 11 to get 3.

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

Calculate 3 to the power of 2 and get 9.

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

The opposite of -3 is 3.

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

Add -3 and 3 to get 0.

$$9+\left(0+3\sqrt{3}\right)^{2}$$

Anything plus zero gives itself.

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

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

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

Calculate 3 to the power of 2 and get 9.

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

The square of \sqrt{3} is 3.

$$9+9\times 3$$

Multiply 9 and 3 to get 27.

$$9+27$$

Add 9 and 27 to get 36.

$$36$$

Factor

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