What is (b^3-8)/(b^2-9)*(b+3)/(b^2+2b+4) ?

The solution to (b^3-8)/(b^2-9)*(b+3)/(b^2+2b+4) is (b-2)/(b-3)

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simplify (b^3-8)/(b^2-9)*(b+3)/(b^2+2b+4)

simplify

\frac{b ^{3} - 8}{b ^{2} - 9} \cdot \frac{b + 3}{b ^{2} + 2 b + 4}

\frac{b - 2}{b - 3}

Solution steps

\frac{b ^{3} - 8}{b ^{2} - 9} \cdot \frac{b + 3}{b ^{2} + 2 b + 4}

Apply the fraction rule:

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

= \frac{( b ^{3} - 8 ) ( b + 3 )}{( b ^{2} - 9 ) ( b ^{2} + 2 b + 4 )}

Factor

b^{3} - 8 : (b - 2) (b^{2} + 2 b + 4)

= \frac{( b - 2 ) ( b ^{2} + 2 b + 4 ) ( b + 3 )}{( b ^{2} - 9 ) ( b ^{2} + 2 b + 4 )}

Cancel the common factor:

b^{2} + 2 b + 4

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

Factor

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

= \frac{( b - 2 ) ( b + 3 )}{( b + 3 ) ( b - 3 )}

Cancel the common factor:

b + 3

= \frac{b - 2}{b - 3}

\frac{x - 9}{8} = 6

8 b^{2} - 10 b + 3

x y + 2 y - x - 2

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

\frac{3}{3}

What is (b^3-8)/(b^2-9)*(b+3)/(b^2+2b+4) ?
The solution to (b^3-8)/(b^2-9)*(b+3)/(b^2+2b+4) is (b-2)/(b-3)