Question 540953
if we replace (x+1) with a, then we get:
{{{a^3*x - 2a^2*x^2 + ax^3}}}
we can factor out an x and an a to get:
{{{ax(a^2 - 2ax + x^2)}}}
we can factor {{{a^2 - 2ax + x^2}}} to get:
{{{ax(a-x)^2}}}
we now replace a with (x+1) to get:
{{{(x+1)*x*((x+1)-x)^2}}}