What is (b^3+1)/(2b+16)*(b^2+8b)/(b^3-b^2+b) ?

The solution to (b^3+1)/(2b+16)*(b^2+8b)/(b^3-b^2+b) is (b+1)/2

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simplify (b^3+1)/(2b+16)*(b^2+8b)/(b^3-b^2+b)

simplify

\frac{b ^{3} + 1}{2 b + 16} \cdot \frac{b ^{2} + 8 b}{b ^{3} - b ^{2} + b}

\frac{b + 1}{2}

Solution steps

\frac{b ^{3} + 1}{2 b + 16} \cdot \frac{b ^{2} + 8 b}{b ^{3} - b ^{2} + b}

Cancel

\frac{b ^{2} + 8 b}{b ^{3} - b ^{2} + b} : \frac{b + 8}{b ^{2} - b + 1}

= \frac{b ^{3} + 1}{2 b + 16} \cdot \frac{b + 8}{b ^{2} - b + 1}

Apply the fraction rule:

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

= \frac{( b ^{3} + 1 ) ( b + 8 )}{( 2 b + 16 ) ( b ^{2} - b + 1 )}

Factor

b^{3} + 1 : (b + 1) (b^{2} - b + 1)

= \frac{( b + 1 ) ( b ^{2} - b + 1 ) ( b + 8 )}{( 2 b + 16 ) ( b ^{2} - b + 1 )}

Cancel the common factor:

b^{2} - b + 1

= \frac{( b + 1 ) ( b + 8 )}{2 b + 16}

Factor

2 b + 16 : 2 (b + 8)

= \frac{( b + 1 ) ( b + 8 )}{2 ( b + 8 )}

Cancel the common factor:

b + 8

= \frac{b + 1}{2}

(a^{2} + 2 b)^{9}

(- 3 + i) (6 + 5 i)

1 \cdot 2 \cdot 3

7^{2} - 4^{2}

16 x^{10}

What is (b^3+1)/(2b+16)*(b^2+8b)/(b^3-b^2+b) ?
The solution to (b^3+1)/(2b+16)*(b^2+8b)/(b^3-b^2+b) is (b+1)/2