#### Answer

$(t+p)(t-p)(t^2-tp+p^2)(t+tp+p^2)$

#### Work Step by Step

The expressions $
t^6
$ and $
p^6
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
t^6-p^6
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(t^3)^2-(p^3)^2
\\\\=
(t^3+p^3)(t^3-p^3)
.\end{array}
The expressions $
t^3
$ and $
p^3
$ are both perfect cubes (the cube root is exact). Hence, $
t^3+p^3
$ is a $\text{
sum
}$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(t^3+p^3)(t^3-p^3)
\\\\=
[(t)^3+(p)^3](t^3-p^3)
\\\\=
(t+p)[(t)^2-t(p)+(p)^2](t^3-p^3)
\\\\=
(t+p)(t^2-tp+p^2)(t^3-p^3)
.\end{array}
The expressions $
t^3
$ and $
p^3
$ are both perfect cubes (the cube root is exact). Hence, $
t^3-p^3
$ is a $\text{
difference
}$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(t+p)(t^2-tp+p^2)(t^3-p^3)
\\\\=
(t+p)(t^2-tp+p^2)[(t)^3-(p)^3]
\\\\=
(t+p)(t^2-tp+p^2)(t-p)[(t)^2+t(p)+(p)^2]
\\\\=
(t+p)(t^2-tp+p^2)(t-p)(t+tp+p^2)
\\\\=
(t+p)(t-p)(t^2-tp+p^2)(t+tp+p^2)
.\end{array}