$\frac{x ^{3} - x}{x ^{2} - 2 x + 1} = \frac{x ^{2} + x}{x - 1}$

Explanation

$\frac{x ^{3} - x}{x ^{2} - 2 x + 1} \overset{◯}{1} \frac{x ^{2} + x}{x - 1}$
①	Simplify $\frac{x ^{3} - x}{x ^{2} - 2 x + 1}$ to $\frac{x ^{2} + x}{x - 1}$ . Factor both the denominator and the numerator, then cancel the common factor. $x - 1$ . $\frac{x ^{3} - x}{x ^{2} - 2 x + 1} = \frac{( x ^{2} + x ) \cdot ( x - 1 )}{( x - 1 ) \cdot ( x - 1 )} = = \frac{x ^{2} + x}{x - 1}$

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$\frac{x ^{3} - x}{x ^{2} - 2 x + 1} = \frac{x ^{2} + x}{x - 1}$