Question 318942
Factor:
{{{t^3+8}}} Notice that this the sum of two cubes:
{{{(t)^3+(2)^2}}}
The sum of two cubes is factored thus:
{{{A^3+B^3 = (A+B)(A^2-AB+B^2)}}} so...
{{{t^3+8 = highlight((t+2)(t^2-2t+4))}}}