Question 62191
Factor:
{{{x^6-(1/729)}}} Do you recognise this as the difference of two cubes?
{{{(x^2)^3 - (1/9)^3}}}
The difference of two cubes can be factored:
{{{A^3 - B^3 = (A-B)(A^2+AB+B^2)}}}
In this problem, A = x^2 and B = 1/9

{{{x^6-(1/729) = (x^2-(1/9))(x^4+x^2/9+1/81)}}} The first parentheses on the right can be factored.
{{{x^6-(1/729) = (x+1/3)(x-1/3)(x^4+x^2/9+1/81)}}}