Prove that the determinant is $(a-b)(b-c)(c-a)(a+b+c)$ I have the determinant : 
\begin{vmatrix}
1 &1 &1 \\
a &b &c \\
a^3 &b^3 &c^3 \\
\end{vmatrix}
How do I prove that this determinant is equal to
$$ (a-b)(b-c)(c-a)(a+b+c) $$

I'll start from where you ended in the comments:
$$bc^3-cb^3-(ac^3-ca^3)+ab^3-ba^3$$
$$bc^3-cb^3-ac^3+ca^3+ab^3-ba^3$$
Now, we need to factor this polynomial. We can do this by trying to group certain terms together and find patterns.


*

*There are two $a^3$ terms: $ca^3-ba^3$. This can be factored into $a^3(c-b)=-a^3(b-c)$

*There are two $a$ terms: $ab^3-ac^3$. This can be factored into $a(b^3-c^3)=a(b-c)(b^2+cb+c^2)$

*There are two terms without an $a$: $bc^3-cb^3$. This can be factored into $bc(c^2-b^2)=bc(c+b)(c-b)=-cb(b+c)(b-c)$


Thus, we can factor out a $b-c$ from all of the terms, giving us:
$$(b-c)(-a^3+a(b^2+cb+c^2)-cb(b-c))=(b-c)(ab^2+abc+ac^2-cb^2-c^2b-a^3)$$
Now, let's try to do what we did before, except now with $b$ instead of $a$.


*

*There are two $b^2$ terms: $ab^2-cb^2$. This can be factored into $b^2(a-c)=-b^2(c-a)$.

*There are two $b$ terms: $abc-c^2b$. This can be factored into $bc(a-c)=-bc(c-a)$.

*There are two terms without a $b$: $ac^2-a^3$. This can be factored into $a(c^2-a^2)=a(c+a)(c-a)$.


Thus, we can factor out a $c-a$, giving us:
$$(b-c)(c-a)(-b^2-bc+a(c+a))=(b-c)(c-a)(ac+a^2-b^2-bc)$$
Now, we can do the same thing we did before, but with $c$:


*

*There are two $c$ terms: $ac-bc$. This can be factored into $c(a-b)$.

*There are two terms without a $c$: $a^2-b^2$. This can be factored into $(a+b)(a-b)$.


Thus, we can factor out an $a-b$, giving us:
$$(a-b)(b-c)(c-a)(c+a+b)=(a-b)(b-c)(c-a)(a+b+c)$$
