Simplify $\frac{1}{(p+q)^3}\left(\frac{1}{p^3}+\frac{1}{q^3}\right)+...$ 
Simplify $\dfrac{1}{(p+q)^3}\big(\dfrac{1}{p^3}+\dfrac{1}{q^3}\big)+\dfrac{3}{(p+q)^4}\big(\dfrac{1}{p^2}+\dfrac{1}{q^2}\big)+\dfrac{6}{(p+q)^5}\big(\dfrac{1}{p}+\dfrac{1}{q}\big)$ if $p\ne -q, p\ne0$ and $q\ne 0$.

$\dfrac{1}{(p+q)^3}\big(\dfrac{1}{p^3}+\dfrac{1}{q^3}\big)+\dfrac{3}{(p+q)^4}\big(\dfrac{1}{p^2}+\dfrac{1}{q^2}\big)+\dfrac{6}{(p+q)^5}\big(\dfrac{1}{p}+\dfrac{1}{q}\big)=\dfrac{p^3+q^3}{p^3q^3(p+q)^3}+\dfrac{3(p^2+q^2)}{p^2q^2(p+q)^4}+\dfrac{6(p+q)}{pq(p+q)^5}$
I guess now it is when I am supposed to calculate the LCD (least common denominator). I have forgotten the algorithm. Can you give me a hint? 
 A: Big hairy guns:
$\dfrac{1}{(p+q)^3}\big(\dfrac{1}{p^3}+\dfrac{1}{q^3}\big)+\dfrac{3}{(p+q)^4}\big(\dfrac{1}{p^2}+\dfrac{1}{q^2}\big)+\dfrac{6}{(p+q)^5}\big(\dfrac{1}{p}+\dfrac{1}{q}\big)=$
$\frac {p^3 + q^3}{p^3q^3(p+q)^3} + 3\frac {p^2 + q^2}{p^2q^2(p+q)^4}+6\frac {p+q}{pq(p+q)^5}=$
$\frac {(p+q)^2(p^3+q^3) + 3(p+q)pq(p^2+q^2) + 6p^2q^2(p+q)}{p^3q^3(p+q)^5}=$
$\frac {(p+q)(p^3+q^3) + 3pq(p^2+q^2) + 6p^2q^2}{p^3q^3(p+q)^4}$
Now $(p+q)^4 = p^4 + 4pq^3 + 6p^2q^2 + 4pq^3 + q^4$ and we seem to have each of those coefficients in the numerator:  $(p+q)(p^3+q^3) =p^4 + q^4 + (p^3q + pq^3)$ and $3qp(p^2 + q^2) =3pq^3 + 3p^3q$ and $6p^2q^2$ is $6p^2q^2$ so
$\frac {(p+q)(p^3+q^3) + 3pq(p^2+q^2) + 6p^2q^2}{p^3q^3(p+q)^4}=$
$\frac {p^4 + 4p^3q + 6p^2q^2 + 4pq^3 +q^4}{p^3q^3(p+q)^4}=$
$\frac {(p+q)^4}{p^3q^3(p+q)^4}=$
$\frac {1}{p^3q^3}$
A: $$
=\dfrac{(p+q)^2}{(p+q)^5} \times \dfrac{p^3 + q^3}{p^3 q^3} + \dfrac{3(p+q)}{(p+q)^5} \times \dfrac{(p^2+q^2)pq}{p^3q^3} + \dfrac{6}{(p+q)^5} \times \dfrac{(p+q)p^2q^2}{p^3q^3}
$$
$$
=\dfrac{(p+q)^2(p^3+q^3)+3(p+q)(p^2+q^2)pq+6(p+q)p^2q^2}{(p+q)^5p^3q^3}
$$
$$
=\dfrac{(p+q)(p^3+q^3)+3(p^2+q^2)pq+6p^2q^2}{(p+q)^4p^3q^3}
$$
numerator $=p^4+p^3q+pq^3+q^4+3q^3q+3pq^3+6p^2q^2=p^4+4p^3q+6p^2q^2+4pq^3+q^4 = (p+q)^4$ 
back to the original:
$=\dfrac{(p+q)^4}{(p+q)^4p^3q^3}$
so, the answer is:
$$
\dfrac{1}{p^3q^3}
$$
