What is (x^2-3x)/(x^2+2x+1)*(x^2-1)/(x^2-9) ?

The solution to (x^2-3x)/(x^2+2x+1)*(x^2-1)/(x^2-9) is (x(x-1))/((x+1)(x+3))

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simplify (x^2-3x)/(x^2+2x+1)*(x^2-1)/(x^2-9)

simplify

\frac{x ^{2} - 3 x}{x ^{2} + 2 x + 1} \cdot \frac{x ^{2} - 1}{x ^{2} - 9}

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

Solution steps

\frac{x ^{2} - 3 x}{x ^{2} + 2 x + 1} \cdot \frac{x ^{2} - 1}{x ^{2} - 9}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{( x ^{2} - 3 x ) ( x ^{2} - 1 )}{( x ^{2} + 2 x + 1 ) ( x ^{2} - 9 )}

Factor

x^{2} - 3 x : x (x - 3)

= \frac{x ( x - 3 ) ( x ^{2} - 1 )}{( x ^{2} + 2 x + 1 ) ( x ^{2} - 9 )}

Factor

x^{2} - 9 : (x + 3) (x - 3)

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

Cancel the common factor:

x - 3

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

Factor

x^{2} - 1 : (x + 1) (x - 1)

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

Factor

x^{2} + 2 x + 1 : (x + 1)^{2}

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

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

(x + 1)^{2} = (x + 1) (x + 1)

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

Cancel the common factor:

x + 1

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

(a + b)^{0}

\frac{1}{x ^{2} + y ^{2} + z ^{2}}

x^{2} + 4 x - 5 x + x^{2}

0 + (- 3)

(7 - 3)^{2}

What is (x^2-3x)/(x^2+2x+1)*(x^2-1)/(x^2-9) ?
The solution to (x^2-3x)/(x^2+2x+1)*(x^2-1)/(x^2-9) is (x(x-1))/((x+1)(x+3))