Does $4(x^3+2x^2+x)^3(3x^2+4x+1) = 4x^3(3x+1)(x+1)^7$?

Question

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...

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

Answer 1

$4(x^3+2x^2+x)^3 . (3x^2+4x+1)$

$4x^3 (x^2+2x+1)^3 (3x^2+4x+1)$

$4x^3 (x+1)^6 (3x^2+4x+1)$

$4x^3 (x+1)^6 (3x+1)(x+1)$

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

So the book resolved your result in more factors.

Answer 2

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.

Answer 3

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 $$ 4x^3 (x + 1)^8 + 8x^4 (x + 1)^7 = 4x^3 (x+1 +2x) (x + 1)^7 = 4x^3 (3x+1) (x + 1)^7 $$

Answer 4

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.

Answer 5

$ \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.