Question 598405
{{{(x-2)^3=x^3-6x^2+12x-8}}}
so I would say it is a cube-shaped box with sides measuring {{{x-2}}}
You may immediately realize that it is the cube of a binomial from the formula
{{{(a+b)^3=4^3+3a^2b+3ab^2+b^3}}} as I did.
Otherwise, you would start factoring, and do a bit more work, but reach the same result. 
To start factoring, you would look for a zero of the polynomial, knowing that the possible rational zeros are:
-1, 1, -2, 2, -4, 4, -8, and 8.
Once you found that 2 is a zero you would be able to factor out {{{(x-2)}}}
and have
{{{V(x)=(x-2)(x^2-4x+4)}}}
The rest of the factoring would be easy.