How can I use the distributive property to rewrite an algebraic fraction? I have an expression:
$$N\left(\dfrac{N(N+1)(N-1)+3N}{3}\right)$$
Can can I use the distrubtive property to form:
$$N^2\left(\dfrac{(N+1)(N-1)+3}{3}\right)$$
If so, how? Could someone advise me on some material so O could strengthen my understanding.
Many Thanks
 A: Both terms in the quantity in the numerator, $N(N+1)(N-1)$ and $3N$, have a factor of N. That means that using the distributive property, you can "pull out" the N. Try that, and see how that might lead you to the final form you're looking for. 
A: The distributive law is about counting: $na+nb=(a+b)n$. In your case you want to count all $N$. In $N(N+1)(N-1)+3N$ how many $N$ are there? Certainly we get $(N+1)(N-1)$ from the first summand and $3$ from the second, adding them yields $(N+1)(N-1)+3$, so we happily achieve $N\bigl((N+1)(N-1)+3\bigr)$.
A: Yes, you can. The distributive law says that $a(b+c)=ab+ac$: given $a(b+c)$, you can multiply it out to get $ab+ac$, and given $ab+ac$, you can factor out the $a$ to get $a(b+c)$. In your case $a=N$, $b=(N+1)(N-1)$, and $c=3$, and you have $$ab+ac=N(N+1)(N-1)+3N$$ in the numerator of the fraction; you may therefore factor out the $N$ to get $$a(b+c)=N\Big((N+1)(N-1)+3\Big)\;.$$ With the denominator of $3$ this is
$$\frac{N\Big((N+1)(N-1)+3\Big)}3=N\left(\frac{(N+1)(N-1)+3}3\right)\;;$$
throw in the extra factor of $N$ outside the fraction, and you have the result.
