<s>[INST] Solve for $x$ in the given determinant. Solve for $x$.  
$$
\begin{vmatrix}  
x^2-a^2&x^2-b^2&x^2-c^2\\  
(x-a)^3&(x-b)^3&(x-c)^3\\  
(x+a)^3&(x+b)^3&(x+c)^3\\  
\end{vmatrix}=0.
$$  
I could factorise each term, but could not find three factors in a single row or column. Atmost, there were $2$. I cannot think of any manipulations of the rows or columns. Please help.
 [/INST] I used brute force to solve this.
When we calculate: 
$$
\begin{vmatrix}  
x^2-a^2&x^2-b^2&x^2-c^2\\  
(x-a)^3&(x-b)^3&(x-c)^3\\  
(x+a)^3&(x+b)^3&(x+c)^3\\  
\end{vmatrix} = 0
$$  
We get:
$$-8 a^3 b^2 x^3+8 a^3 c^2 x^3+8 a^2 b^3 x^3+24 a^2 b x^5-8 a^2 c^3 x^3-24 a^2 c x^5-24 a b^2 x^5+24 a c^2 x^5-8 b^3 c^2 x^3+8 b^2 c^3 x^3+24 b^2 c x^5-24 b c^2 x^5 = 0$$
We see that we can divide by $-8 x^3$ and eliminate a triple root of $x = 0$, which reduces to:
$$(a-b) (a-c) (b-c) \left(-3x^2 + a (b+c)+b c\right) = 0$$
Using the quadratic, we find the other two roots as:
$$x=\pm ~\dfrac{\sqrt{a b+a c+b c}}{\sqrt{3}}$$
</s>