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yeongil · #4 June 27th, 2009, 04:22 PM

2)
$\frac{a^4}{2a^2 - 4a^4} \cdot \frac{8a^3 + 1}{2a + 4a^2}$

Begin by factoring (note that the numerator of the 2nd fraction is a sum of 2 cubes):
$= \frac{a^4}{-2a^2(2a^2 - 1)} \cdot \frac{(2a + 1)(4a^2 - 2a + 1)}{2a(2a + 1)}$

Cancel things out:
$\begin{aligned}&= \frac{a^4}{-2a^2(2a^2 - 1)} \cdot \frac{{\color{red}(2a + 1)}(4a^2 - 2a + 1)}{2a{\color{red}(2a + 1)}} \\&= \frac{{\color{red}a^4}}{-2{\color{red}a^2}(2a^2 - 1)} \cdot \frac{4a^2 - 2a + 1}{2{\color{red}a}} \\&= \frac{a}{-2(2a^2 - 1)} \cdot \frac{4a^2 - 2a + 1}{2} \\\end{aligned}$

Multiply:
$= -\frac{a(4a^2 - 2a + 1)}{4(2a^2 - 1)}$

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