Solve (x^2+2x/sqrt{3}+1/3)(x^2-2x/sqrt{3}+1/3)

$$( x ^ { 2 } + \frac { 2 x } { \sqrt { 3 } } + \frac { 1 } { 3 } ) ( x ^ { 2 } - \frac { 2 x } { \sqrt { 3 } } + \frac { 1 } { 3 } )$$

Evaluate

$$\frac{\left(3x^{2}-1\right)^{2}}{9}$$

Short Solution Steps

Rationalize the denominator of \frac{2x}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\left(x^{2}+\frac{2x\sqrt{3}}{\left(\sqrt{3}\right)^{2}}+\frac{1}{3}\right)\left(x^{2}-\frac{2x}{\sqrt{3}}+\frac{1}{3}\right)$$

The square of \sqrt{3} is 3.

$$\left(x^{2}+\frac{2x\sqrt{3}}{3}+\frac{1}{3}\right)\left(x^{2}-\frac{2x}{\sqrt{3}}+\frac{1}{3}\right)$$

Since \frac{2x\sqrt{3}}{3} and \frac{1}{3} have the same denominator, add them by adding their numerators.

$$\left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)\left(x^{2}-\frac{2x}{\sqrt{3}}+\frac{1}{3}\right)$$

Rationalize the denominator of \frac{2x}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)\left(x^{2}-\frac{2x\sqrt{3}}{\left(\sqrt{3}\right)^{2}}+\frac{1}{3}\right)$$

The square of \sqrt{3} is 3.

$$\left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)\left(x^{2}-\frac{2x\sqrt{3}}{3}+\frac{1}{3}\right)$$

Since \frac{2x\sqrt{3}}{3} and \frac{1}{3} have the same denominator, add them by adding their numerators.

$$\left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)\left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)$$

Multiply x^{2}+\frac{2x\sqrt{3}+1}{3} and x^{2}+\frac{2x\sqrt{3}+1}{3} to get \left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)^{2}.

$$\left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2} times \frac{3}{3}.

$$\left(\frac{3x^{2}}{3}+\frac{2x\sqrt{3}+1}{3}\right)^{2}$$

Since \frac{3x^{2}}{3} and \frac{2x\sqrt{3}+1}{3} have the same denominator, add them by adding their numerators.

$$\left(\frac{3x^{2}+2x\sqrt{3}+1}{3}\right)^{2}$$

To raise \frac{3x^{2}+2x\sqrt{3}+1}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3x^{2}+2x\sqrt{3}+1\right)^{2}}{3^{2}}$$

Square 3x^{2}+2x\sqrt{3}+1.

$$\frac{9x^{4}+12\sqrt{3}x^{3}+4\left(\sqrt{3}\right)^{2}x^{2}+6x^{2}+4\sqrt{3}x+1}{3^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{9x^{4}+12\sqrt{3}x^{3}+4\times 3x^{2}+6x^{2}+4\sqrt{3}x+1}{3^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{9x^{4}+12\sqrt{3}x^{3}+12x^{2}+6x^{2}+4\sqrt{3}x+1}{3^{2}}$$

Combine 12x^{2} and 6x^{2} to get 18x^{2}.

$$\frac{9x^{4}+12\sqrt{3}x^{3}+18x^{2}+4\sqrt{3}x+1}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{9x^{4}+12\sqrt{3}x^{3}+18x^{2}+4\sqrt{3}x+1}{9}$$

Factor

$$\frac{\left(3x^{2}-1\right)^{2}}{9}$$