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: $4x^3(3x+1)(x+1)^7$

$ \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.  
