Simplify (x-5)/(x-3)*(x^5-x^3)/(x^2-6x+5)

$\frac{x - 5}{x - 3} \cdot \frac{x^{5} - x^{3}}{x^{2} - 6 x + 5}$

Simplify the numerator.

Factor $x^{3}$ out of $x^{5} - x^{3}$ .

Factor $x^{3}$ out of $x^{5}$ .

$\frac{x - 5}{x - 3} \cdot \frac{x^{3} x^{2} - x^{3}}{x^{2} - 6 x + 5}$

Factor $x^{3}$ out of $- x^{3}$ .

$\frac{x - 5}{x - 3} \cdot \frac{x^{3} x^{2} + x^{3} \cdot - 1}{x^{2} - 6 x + 5}$

Factor $x^{3}$ out of $x^{3} x^{2} + x^{3} \cdot - 1$ .

$\frac{x - 5}{x - 3} \cdot \frac{x^{3} (x^{2} - 1)}{x^{2} - 6 x + 5}$

Rewrite $1$ as $1^{2}$ .

$\frac{x - 5}{x - 3} \cdot \frac{x^{3} (x^{2} - 1^{2})}{x^{2} - 6 x + 5}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$\frac{x - 5}{x - 3} \cdot \frac{x^{3} (x + 1) (x - 1)}{x^{2} - 6 x + 5}$

Factor $x^{2} - 6 x + 5$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $5$ and whose sum is $- 6$ .

$- 5, - 1$

Write the factored form using these integers.

$\frac{x - 5}{x - 3} \cdot \frac{x^{3} (x + 1) (x - 1)}{(x - 5) (x - 1)}$

Simplify terms.

Cancel the common factor of $x - 5$ .

Cancel the common factor.

$\frac{x - 5}{x - 3} \cdot \frac{x^{3} (x + 1) (x - 1)}{(x - 5) (x - 1)}$

Rewrite the expression.

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

Multiply $\frac{1}{x - 3}$ by $\frac{x^{3} (x + 1) (x - 1)}{x - 1}$ .

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

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify (x-5)/(x-3)*(x^5-x^3)/(x^2-6x+5)

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