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.
 A: 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*}
A: \begin{gathered}
\left|\begin{array}{ccc}
b c & a c & a b \\
a b+a c & a b+b c & a c+b c \\
a^{2} & b^{2} & c^{2}
\end{array}\right|=b c \cdot(a b+b c) \cdot c^{2}+a c \cdot(a c+b c) \cdot a^{2}+a b \cdot(a b+a c) \cdot b^{2}-a^{2} \cdot(a b+b c) \cdot(a b)-b^{2} \cdot(a c+b c) \cdot(b c)-c^{2} \cdot(a b+a c) \cdot(a c) \\
\equiv \\
=a^{2} b^{4}-a^{4} b^{2}-a^{2} c^{4}+b^{2} c^{4}+a b^{2} c^{3}-a^{2} b c^{3}+a^{4} c^{2}-b^{4} c^{2}-a b^{3} c^{2}+a^{3} b c^{2}+a^{2} b^{3} c-a^{3} b^{2} c
\end{gathered}
A: Adding the first row to the second row makes the second row constant which implies
$$\Delta=(ab+bc+ca)\begin{vmatrix}bc&ca&ab\\1&1&1\\a^2&b^2&c^2\end{vmatrix}. \tag{1}$$
Notice that the remaining determinant is alternating. That is, for
example, exchanging $\,a\,$ for $\,b\,$ is the same as switching
the first two columns and hence changes its sign. Thus it is divisible
by $\,(a-b)(b-c)(c-a)\,$ while the remaining factor is of degree one,
is symmetric and therefore must be $\,(a+b+c)\,$times a constant which can be found to be $1$.
An alternative way to get this is from the original determinant adding the
third row to the second row gives it a common linear factor of $\,(a+b+c).\,$ Thus
$$\Delta=(ab+bc+ca)(a-b)(b-c)(c-a)(a+b+c). \tag{2}$$
