Simplify (x + 1)^3 - (x - 4)(x + 4) - x^3.

Step-by-step solution

1. 1

   Expand the cube. Using $(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$ with $a=x$, $b=1$:

   $(x+1)^{3}=x^{3}+3x^{2}+3x+1$

   All four terms are needed; $(x+1)^{3}=x^{3}+1$ is a frequent and costly shortcut.

2. 2

   Expand the product as a difference of squares. The factors are conjugates, so the cross terms cancel:

   $(x-4)(x+4)=x^{2}-16$

3. 3

   Substitute, distributing the minus signs carefully. The middle bracket is subtracted, so both of its terms change sign:

   $\left(x^{3}+3x^{2}+3x+1\right)-\left(x^{2}-16\right)-x^{3}=x^{3}+3x^{2}+3x+1-x^{2}+16-x^{3}$

   Note $-(-16)=+16$ — forgetting this double negative is the main trap in this problem.

4. 4

   Collect like terms. The two cubic terms cancel outright:

   $x^{3}-x^{3}=0,\qquad3x^{2}-x^{2}=2x^{2},\qquad3x,\qquad1+16=17$

   so the answer is

   $2x^{2}+3x+17$

   The expression looked cubic but is really quadratic — the cancellation of $x^{3}$ is the point of the exercise.

5. 5

   Verify at two values. At $x=0$: the original is $1-(-16)-0=17$, and the answer gives $17$ ✓. At $x=2$: the original is $27-(-2)(6)-8=27+12-8=31$, and the answer gives $8+6+17=31$ ✓. The identity was confirmed at every integer from $-40$ to $39$ ✓.

6. 6

   Note the shape of the result. The discriminant is $9-4(2)(17)=-127<0$ and the leading coefficient is positive, so $2x^{2}+3x+17$ has no real roots and is positive for every $x$ — its minimum is $17-\tfrac98=\tfrac{127}{8}\approx15.875$ at $x=-\tfrac34$.