Question 212746
Factor completely:
{{{3x^4-81x}}} First, factor out 3x from each term.
{{{3x(x^3-27)}}} Notice that the binomial in the parentheses is the difference of two cubes.
{{{3x((x)^3-(3)^3)}}}
The difference of two cubes is factorable thus:
{{{A^3-B^3 = (A-B)(A^2+AB+B^2)}}}, so...
{{{3x((x)^3-(3)^3) = highlight(3x(x-3)(x^2+3x+9))}}}