If \(\frac{1}{p}\) = \(\frac{a^2 + 2ab + b^2}{a – b}\) and \(\frac{1}{q}\) = \(\frac{a + b}{a^2 – 2ab + b^2}\) Find \(\frac{p}{q}\)
- \(\frac{a + b}{a - b}\)
-
\(\frac{1}{a^2 - b^2}\)
- \(\frac{a - b}{a + b}\)
- a2 - b2
The correct answer is: B
Explanation
\(\frac{1}{p} = \frac{a^{2} + 2ab + b^{2}}{a - b}\)
\(\frac{1}{q} = \frac{a + b}{a^{2} - 2ab + b^{2}}\)
\(\frac{1}{p} = \frac{(a + b)^{2}}{a - b}\)
\(\frac{1}{q} = \frac{a + b}{(a - b)^{2}}\)
\(\therefore p = \frac{a - b}{(a + b)^{2}}\)
\(\frac{p}{q} = p \times \frac{1}{q} = \frac{a - b}{(a + b)^{2}} \times \frac{a + b}{(a - b)^{2}}\)
= \(\frac{1}{(a + b)(a - b)}\)
= \(\frac{1}{a^{2} - b^{2}}\)