#### Answer

$10(c+1)(c^2-c+1)(c-1)(c^2+c+1)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Get the $GCF$ of the given expression, $
10c^6-10
.$ Then use the factoring of the difference of $2$ squares and the factoring of the sum and difference of $2$ cubes.
$\bf{\text{Solution Details:}}$
The $GCF$ of the terms is $
10
$ since it is the highest number that can evenly divide (no remainder) all the given terms. Factoring the $GCF,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
10c^6-10
\\\\=
10(c^6-1)
.\end{array}
The expressions $
c^6
$ and $
1
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
c^6-1
$ 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}
10(c^6-1)
\\\\=
10(c^3+1)(c^3-1)
.\end{array}
The expressions $
c^3
$ and $
1
$ are both perfect cubes (the cube root is exact). Hence, $
c^3+1
$ 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}
10(c^3+1)(c^3-1)
\\\\=
10[(c)^3+(1)^3](c^3-1)
\\\\=
10(c+1)[(c)^2-c(1)+(1)^2](c^3-1)
\\\\=
10(c+1)(c^2-c+1)(c^3-1)
.\end{array}
The expressions $
c^3
$ and $
1
$ are both perfect cubes (the cube root is exact). Hence, $
c^3-1
$ 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}
10(c+1)(c^2-c+1)(c^3-1)
\\\\=
10(c+1)(c^2-c+1)[(c)^3-(1)^3]
\\\\=
10(c+1)(c^2-c+1)(c-1)[(c)^2+c(1)+(1)^2]
\\\\=
10(c+1)(c^2-c+1)(c-1)(c^2+c+1)
.\end{array}