I have a math question I'm really confused on from my teacher.
The question is based on consecutive integers.
Ex. $1\times2\times3+2= 8$ which is equivalent to $2^3$ ($2\times2\times2$)
The question is that if the same applies for $x\times(x+1)\times(x+2)+x+1$, which the answer should be equal to $(x+1)^3$, what is the formula to prove this?
Thank you!
Expand the expressions "by hand":
$$x(x+1)(x+2)+x+1=x(x^2+2x+x+2)+x+1=x^3+3x^2+3x+1,$$
$$(x+1)^3=(x+1)(x^2+2x+1)=(x^3+2x^2+x)+(x^2+2x+1)=x^3+3x^2+3x+1.$$
The last equality is also known to be a remarkable identity ($(a+b)^3$) or can be retrieved from the Binomial formula.
Alternatively, as these expressions are polynomials of the third degree, you can also prove equivalence by testing with four different values.
$$\begin{align}(-2)(-2+1)(-2+2)+(-2)+1=-1,\ &(-2+1)^3=-1,\\ (-1)(-1+1)(-1+2)+(-1)+1=0,\ &(-1+1)^3=0,\\ 0(0+1)(0+2)+0+1=1,\ &(0+1)^3=1,\\ 1(1+1)(1+2)+1+1=8,\ &(1+1)^3=8.\end{align}$$
This is valid as a proof (negative arguments were chosen to keep small values).
By a little of formula massaging,
$$x(x+1)(x+2)+x+1=x(x+1)(x+2)+(x+1)=(x+1)(x(x+1+1)+1)=(x+1)(x(x+1)+x+1)=(x+1)(x+1)(x+1).$$
If you prefer, set $t=x+1$, and
$$(t-1)t(t+1)+t=t(t^2-1)+t=t^3.$$
Expanding $(x+1)^3$ gives us $x^3+3x^2+3x+1$. Expanding $x(x+1)(x+2)$ gives us $x^3+3x^2+2x$. So $x(x+1)(x+2)+x+1=x^3+3x^2+2x+x+1=x^3+3x^2+3x+1$.
