find the second derivative $(x^3+x-1)(x^3+1)$ $$(x^3+x-1)(x^3+1)$$
$$=x^6+x^4-x^3+x^3+x-1$$
$$f'(x)= 6x^5+4x^3-3x^2+3x^2+1$$
Am I suppose to cancel out $-3x^2+3x^2$?
$$f''(x)= 30x^4+12x^2-6x+6x$$
Can someone check my work please? Thank you so much!
 A: You could also work it out this way:
The Product Rule gives us $ \ [ fg ]' \ = \ f' \cdot g \ + \ f \cdot g' $ .  Applying it again gives us
$$ [fg]'' \ = \ f'' \cdot g \ + \ 2 \cdot f' \cdot g' \ + \ f \cdot g'' \ . $$ 
(The "higher-derivative" Product Rule has a resemblance to the Binomial Theorem. [It's more like a relation to the Theorem if you use derivative operators.])
For the functions $ \ f(x) \ = \ x^3 \ + \ x \ - \ 1 \ \ $ and $ \ g(x) \ = \ x^3 \ +  \ 1 \ \ , $ we have
$$ f'(x) \ = \ 3x^2 \ + \ 1 \ \ , \ \  f''(x) \ = \ 6x \ \ , \ \ g'(x) \ = \ 3x^2 \ \ , \ \ g''(x) \ = \ 6x \ \ ,$$
making the second derivative of the product $ \ fg \ $
$$ 6x \ \cdot \ (x^3 + 1 ) \ + \ 2 \ \cdot \ (3x^2 + 1) \ \cdot \ 3x^2 \ + \ (x^3 \ + \ x \ - \ 1 ) \ \cdot \ 6x $$
$$ = \ 6x^4 \ + \ 6x \ + \ 18x^4 \ + \ 6x^2 \ + \ 6x^4 \ + \ 6x^2 \ - \ 6x $$
$$ = \ 30x^4 \ + \ 12x^2 \ \ . $$
Granted, it's not much of a saving in calculation effort here, but it can be handy for  more complicated functions...
A: You can cancel the 6x, but yes, that is correct according to wolfram alpha.
