Solve x(x-1)-(x-1/3)(x+1/3)

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

Evaluate

$$\frac{1}{9}-x$$

Solution Steps

Use the distributive property to multiply x by x-1.

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

Consider \left(x-\frac{1}{3}\right)\left(x+\frac{1}{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate \frac{1}{3} to the power of 2 and get \frac{1}{9}.

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

To find the opposite of x^{2}-\frac{1}{9}, find the opposite of each term.

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

The opposite of -\frac{1}{9} is \frac{1}{9}.

$$x^{2}-x-x^{2}+\frac{1}{9}$$

Combine x^{2} and -x^{2} to get 0.

$$-x+\frac{1}{9}$$

Expand

$$\frac{1}{9}-x$$

Solution Steps

Use the distributive property to multiply x by x-1.

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

Consider \left(x-\frac{1}{3}\right)\left(x+\frac{1}{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate \frac{1}{3} to the power of 2 and get \frac{1}{9}.

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

To find the opposite of x^{2}-\frac{1}{9}, find the opposite of each term.

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

The opposite of -\frac{1}{9} is \frac{1}{9}.

$$x^{2}-x-x^{2}+\frac{1}{9}$$

Combine x^{2} and -x^{2} to get 0.

$$-x+\frac{1}{9}$$