How do you reduce to lowest term $\frac{x^{3} - 3 x}{x^{2} - 3 x}$ $(x^3-3x)/(x^2-3x)$ ?

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Answer 1 · 2016-11-15T15:21:16

$\frac{(x^{2} - 3)}{(x - 3)}$ $((x^2 -3))/((x-3)$

$x \neq 0 and x \neq 3$ $x !=0 and x!=3$

Find factors of the numerator and denominator

$\frac{x^{3} - 3 x}{x^{2} - 3 x}$ $(x^3-3x)/(x^2-3x)$

= $\frac{x (x^{2} - 3)}{x (x - 3)} \leftarrow x \neq 0 and x \neq 3$ $(x(x^2 -3))/(x(x-3))" "larr x !=0 and x!=3$

= $\frac{(x^{2} - 3)}{(x - 3)}$ $((x^2 -3))/((x-3)$

How do you reduce to lowest term x3−3xx2−3x(x^3-3x)/(x^2-3x)?