Does $4(x^3+2x^2+x)^3(3x^2+4x+1) = 4x^3(3x+1)(x+1)^7$? Please provide proof
The answer in the back of the book is different to both my calculations and also the online calculator I crosschecked my answer with...

The Question:
"Differentiate with respect to $x$:"
$
(x^3+2x^2+x)^4
$
My Answer: $4(x^3+2x^2+x)^3(3x^2+4x+1)$
The Book's Answer: $4x^3(3x+1)(x+1)^7$
 A: You're both right.
Note that
$$
x^3+2x^2+x = x (x + 1)^2
$$
$$
3x^2+4x+1 = (3 x + 1) (x + 1)
$$
On the other hand, you might have started with
$$
(x^3+2x^2+x)^4 = x^4 (x + 1)^8
$$
whose derivative is
$$
4x^3 (x + 1)^8 + 8x^4 (x + 1)^7
= 4x^3 (x+1 +2x) (x + 1)^7
= 4x^3 (3x+1) (x + 1)^7
$$
A: Note that $(x^3+2x^2+x)=x(x^2+2x+1)=x(x+1)^2$.
Also, $3x^2+4x+1=(x+1)(3x+1)$
Apply those factorizations to your answer, and collect like factors together; you should get the book's answer.
A: Since $(\forall x\in\mathbb{R}):4(x^3+2x^2+x)^3(3x^2+4x+1)=4x^3(3x+1)(x+1)^7$, both answers are correct. But your answer is more natural.
A: $4(x^3+2x^2+x)^3 . (3x^2+4x+1)$
$4x^3 (x^2+2x+1)^3 (3x^2+4x+1)$
$4x^3 (x+1)^6      (3x^2+4x+1)$
$4x^3 (x+1)^6      (3x+1)(x+1)$
$4x^3 (x+1)^7 (3x+1)$
So the book resolved your result in more factors.
A: $ \frac{d}{dx}(x^3+2x^2+x)^4 = 4(x^3+2x^2+x)^3\cdot(\frac{d}{dx}(x^3+2x^2+x))$ by the chain rule, then we get: $ \frac{d}{dx}(x^3+2x^2+x)^4 = 4(x^3+2x^2+x)^3\cdot(3x^2+4x+1) = 4x^3(x^2+2x+1)^3\cdot(3x^2+4x+1) $.
Now the rest should just be algebra: (note that $(x+1)^2=(x^2+2x+1)$)
$ 4x^3((x+1)^2)^3(3x^2+4x+1) = 4x^3(x+1)^6(3x^2+4x+1) $,
then show that $ (x+1)(3x+1)= (3x^2 +4x + 1) $, and insert in the above equation:
$ \frac{d}{dx}(x^3+2x^2+x)^4 = 4x^3(x+1)^7(3x+1) $, and you're done.  
