Simplifying radical expression $\frac{\sqrt{p-4q}}{\sqrt{3q^2+4pq+p^2}}\cdot\frac{\sqrt{p+3q}}{\sqrt{p^2+6pq+8q^2}}\div\frac1{\sqrt{p^2+3pq+2q^2}}$

Question

Problem:

$$\dfrac{\sqrt{p-4q}}{\sqrt{3q^2+4pq+p^2}}\cdot\dfrac{\sqrt{p+3q}}{\sqrt{p^2+6pq+8q^2}}\div\dfrac1{\sqrt{p^2+3pq+2q^2}}$$

My try:

$$\dfrac{\sqrt{p-4q}}{\sqrt{3p+q}\sqrt{p+q}}\cdot\dfrac{\sqrt{p+3q}}{\sqrt{p+4q}\sqrt{p+2q}}\cdot\dfrac{\sqrt{p+2q}\sqrt{p+q}}1=$$ $$\dfrac{\sqrt{p-4q}\,\sqrt{p+3q}}{\sqrt{3p+q}\,\sqrt{p+4q}}=$$ $$\dfrac{\sqrt{p^2-pq-12q}}{\sqrt{3p^2+13pq+4q^2}}$$

How do I continue?

Answer 1

Based on the problem as you've (and I've) corrected it, you failed to mind your $p$s and $q$s. It should have been

$$\frac{\sqrt{p-4q}}{\sqrt{\color{red}{3q+p}}\sqrt{p+q}}\cdot \frac{\sqrt{p+3q}}{\sqrt{p+4q}\sqrt{p+2q}}\cdot \frac{\sqrt{p+2q}\sqrt{p+q}}1$$ $$=\frac{\sqrt{p-4q}}{\sqrt{p+4q}}$$ $$=\frac{\sqrt{p-4q}\sqrt{p+4q}}{\sqrt{p+4q}\sqrt{p+4q}}$$ $$=\frac{\sqrt{p^2-16q^2}}{|p+4q|}.$$