Answer
(t+2)(t$^2$-2t+t$^2$)(t-2)(t$^2$+2t+t$^2$)
Work Step by Step
t$^6$-64
This expression is a difference of squares. If a is the square root of the first term and b is the square root of the second term, the factorization is (a+b)(a-b)
$\sqrt {t^6}$=t$^3$ and
$\sqrt {64}$=8. So the factorization is
(t$^3$+8)(t$^3$-8).
Now these two polynomials are a sum of cubes and a difference of cubes. If the cube root of the first term in each polynomial is a and the cube root of the second term in each polynomial is b, then the full factorization of both is (a+b)(a$^2$-ab+b$^2$)(a-b)(a$^2$+ab+b$^2$)
$\sqrt[3] {t^3}$=t and
$\sqrt[3] {8}$=2. So the factorization is
(t+2)(t$^2$-2t+t$^2$)(t-2)(t$^2$+2t+t$^2$).
