Without expanding, prove that abc can be factored out

Question

Without expanding, prove that

$$\begin{vmatrix}bc&c^2&b^2\\c^2&ca&a^2\\b^2&a^2&ab\end{vmatrix}=abc\begin{vmatrix}a&c&b\\c&b&a\\b&a&c\end{vmatrix}$$

How should my thinking be when I see a matrix equal its transpose like this?

Answer 1

One can multiply:
$1$st row by $a$,
$2$nd row by $b$,
$3$rd row by $c$,
and (of course) divide determinant by $abc$ after that: $$ \left| \begin{array}{ccc} bc & c^2 & b^2 \\ c^2 & ca & a^2 \\ b^2 & a^2 & ab \end{array} \right| = \dfrac{1}{abc}\left| \begin{array}{ccc} abc & ac^2 & ab^2 \\ bc^2 & abc & a^2b \\ b^2c & a^2c & abc \end{array} \right|.\tag{1} $$

Then divide:
$1$st column by $bc$,
$2$nd column by $ac$,
$3$rd column by $ab$,
and (of course) multiply whole result by $bc\cdot ac\cdot ab = a^2b^2c^2$:

$$ (1) = \dfrac{1}{abc}\cdot a^2b^2c^2 \left| \begin{array}{ccc} a & c & b \\ c & b & a \\ b & a & c \end{array} \right| = abc \left| \begin{array}{ccc} a & c & b \\ c & b & a \\ b & a & c \end{array} \right| .\tag{2} $$

---

Note: this manipulation is correct when $a\ne 0$, $b\ne 0$, $c\ne 0$. If one of them is zero, then LHS-determinant is equal to $0$ (starting formula is correct too in this case).