Calculate: $\Delta=\left|\begin{array}{ccc} b c & c a & a b \\ a(b+c) & b(c+a) & c(a+b) \\ a^{2} & b^{2} & c^{2} \end{array}\right|$ Calculate:
$$\Delta=\left|\begin{array}{ccc}
b c & c a & a b \\
a(b+c) & b(c+a) & c(a+b) \\
a^{2} & b^{2} & c^{2}
\end{array}\right|$$
Does anyone know any easy way to calculate this determinant?
I tried the classic way but I was wondering if anyone knows an easier method.

If you add the first row to the second, the second row becomes three copies of $ab+bc+ca$, and you can write
$$\Delta=(ab+bc+ca)\begin{vmatrix}bc&ca&ab\\1&1&1\\a^2&b^2&c^2\end{vmatrix}.$$
Now, subtract the first column from the second and third to get
$$\Delta=(ab+bc+ca)\begin{vmatrix}bc&c(a-b)&b(a-c)\\1&0&0\\a^2&b^2-a^2&c^2-a^2\end{vmatrix};$$
you can then factor out $(a-b)(a-c)$ to get
$$\Delta=(ab+bc+ca)(a-b)(a-c)\begin{vmatrix}bc&c&b\\1&0&0\\a^2&-a-b&-a-c\end{vmatrix}.$$
Expanding by minors along the second row gives
\begin{align*}
\Delta&=(ab+bc+ca)(a-b)(a-c)(c(a+c)-b(a+b))\\
&=(ab+bc+ca)(a-b)(b-c)(c-a)(a+b+c).
\end{align*}
