Solving a Polynomial Equation by Factoring. $n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.
 A: Hint: remember the binomial theorem $(a+b)^3 = a^3 + 3 a^2 b + 3 b^2 a + b^3.$
Now guess what $a$ and $b$ might be here.
A: Hint: \begin{align} n^3+12n^2+48n+64&= (n^3+64)+(12n^2+48n)\\  &= (n^3+4^3)+12n(n+4)\\
&=... \end{align}
A: Here it is an approach based on the Binomial Theorem:
\begin{align*}
n^{3} + 12n^{2} + 48n + 64 = {3\choose 0}n^{3}4^{0} + {3\choose 1}n^{2}4^{1} + {3\choose 2}n4^{2} + {3\choose 3}n^{0}4^{3} = (n+4)^{3}
\end{align*}
A: If $$P(n)=(n-a)^3$$
then $ a $ is a root of
$$P''(n)=0$$
with
$$P'(n)=3n^2+24n+48$$
and
$$P''(n)=6n+24$$
thus $$a=-4 \;;\;P(n)=(n+4)^3$$
A: Differentiate $f(n) = n^3+12n^2+48n+64$ to get $f'(n) = 3n^2 + 24n + 48 = 3(n^2 + 8n + 16) = 3(n+4)^2$.
Hence $\int f'(n) \ dn = 3 \int (n+4)^2 \ dn = 3 \cdot \left(\frac{(n+4)^3}{3} + C \right) = (n+4)^3 + C' = f(n)$, by the fundamental theorem of calculus. Since $f(-4) = C' \Rightarrow 0 = C'$, $f(n) = (n+4)^3$.
A: Factor out a $64$ to get:
$$64 \left(\frac{n^3}{64} + \frac{12n^2}{64} + \frac{48n}{64} +1\right) = 64 \left(\frac{n^3}{64} + 3 \cdot\frac{n^2}{16} + 3 \cdot \frac{n}{4} +1\right)$$
$$= 64 \left( \left(\frac{n}{4}\right)^3 + 3\left(\frac{n}{4}\right)^2 + 3 \left(\frac{n}{4}\right) + 1 \right)$$
$$= 64(n/4 + 1)^3 = (4 \cdot n/4 + 4 \cdot1)^3 = (n+4)^3$$
where we have used the binomial theorem in the last line, and the fact that $4^3 = 64$.
This can be seen as the reverse of the AC method, where we instead factor out the largest perfect square of the constant term. This is $64 = 8^2$ in this question.
A: You were going in the correct direction, as you said:
$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
If we apply the same here:
$^3+12^2+48+64 = ^3+ 4^3 + 12^2+48 = (n+4)(n^2 - 4n + 16) + 12n(n+4)$
If we take (n+4) common, we get:
$(n+4)(n^2 - 4n + 16 + 12n) = (n+4)(n^2 + 8n + 16) = (n+4)(n+4)^2 = (n+4)^3$Cheers
A: You can simplify expression to see what happens.
For example, you see that $12,48,64 $ have a common divisor: $4$
If you accept $4=n$, then you get
$$m^3+3m^2n+3mn^2+n^3=(m+n)^3.$$
