Let $a,b$ and $c$ be real numbers. Evaluate the following determinant:
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}$$
after long calculation I get that the answer will be $0$. Is there any short processs? Please help someone thank you.
If $b=0,$ $$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix} =\begin{vmatrix}0 &0&c\cr c^2a^2&ca&c+a\cr 0&0&a\cr\end{vmatrix}$$
Now, if $a=0,$ $$\text{the determinant becomes }\begin{vmatrix}0 &0&c\cr 0&0&c\cr 0&0&0\cr\end{vmatrix}=0$$
else for $ca\ne 0$ $$\text{the determinant becomes }c^3a^3\begin{vmatrix}0 &0&c\cr 1&1&c\cr 0&0&0\cr\end{vmatrix}=0$$
So, if $abc\ne 0,$
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}$$
$$=\frac1{abc}\begin{vmatrix}a(b^2c^2) &abc& a(b+c)\cr b(c^2a^2)&cab&b(c+a)\cr c(a^2b^2)&abc&c(a+b)\cr\end{vmatrix} \text{ applying } R_1'=aR_1, R_2'=bR_2, R_3'=cR_3$$
$$=abc\begin{vmatrix}bc &1& a(b+c)\cr ca&1&b(c+a)\cr ab&1&c(a+b)\cr\end{vmatrix}\text{ applying } C_1'=\frac{C_1}{abc} \text{ and } C_2'=\frac{C_2}{abc}$$.
$$=abc\begin{vmatrix}bc &1& a(b+c)+bc\cr ca&1&b(c+a)+ca\cr ab&1&c(a+b)+ab\cr\end{vmatrix} \text{ applying } C_3'=C_3+C_1$$
If $ab+bc+ca=0,$ $$\text{the determinant becomes } abc\begin{vmatrix}bc &1& 0\cr ca&1&0\cr ab&1&0\cr\end{vmatrix}=0 $$
Else $$=abc(ab+bc+ca)\begin{vmatrix}bc &1& 1\cr ca&1&1\cr ab&1&1\cr\end{vmatrix} \text{ applying } C_3'=\frac{C_3}{ab+bc+ca}$$
$$=abc(ab+bc+ca)\cdot 0 \text { as the 2nd & the last columns are identical.}$$
Expansion on column 1 isn't too bad. In each cofactor the terms $abc$ cancel. It becomes
$$b^2c^2 (ca^2-a^2b) +c^2a^2 (ab^2-b^2c)+a^2b^2(bc^2-c^2a)$$ which is $a^2b^2c^2[(c-b)+(a-c)+(b-a)]=0.$
We can assume $\,abc\neq 0\,$ , otherwise the determinant is zero at once (why? For example, suppose $\,b=0\,$ . Then either $\,ac=0\,$ and we get a row of zeros, or the 3rd row is a multiple of the 1st one...and likewise if $\,a=0\vee c=0\,$ ):
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}\stackrel{aR_1\,,\,bR_2\,,\,cR_3}{\longrightarrow}\begin{vmatrix}ab^2c^2 &abc& ab+ac\cr a^2bc^2&abc&ab+bc\cr a^2b^2c&abc&ac+bc\cr\end{vmatrix}\stackrel{R_2-R_1\,,\,R_3-R_1}\longrightarrow$$
$$\begin{vmatrix}ab^2c^2 &abc& ab+ac\cr abc^2(a-b)&0&-c(a-b)\cr ab^2c(a-c)&0&-b(a-c)\cr\end{vmatrix}\stackrel{\text{develop by 2nd column}}\longrightarrow$$
$$-abc\begin{vmatrix}abc^2(a-b)&-c(a-b)\\ab^2c(a-c)&-b(a-c)\end{vmatrix}$$
And now factor out $\,c(a-b)\,\,,\,\,b(a-c)\,$ from first and second row resp. , and get
$$-ab^2c^2(a-b)(a-c)\begin{vmatrix}abc&-1\\abc&-1\end{vmatrix}=0$$
Finally, pay attention to the fact that the original determinant is zero (or not) without any regard to the above operations
\begin{align*} \begin{vmatrix}b^2c^2&bc&b+c\\c^2a^2&ca&c+a\\a^2b^2&ab&a+b\end{vmatrix} \stackrel{\large R_1-R_3\,,\,R_2-R_3}{\longrightarrow} &(c-a)(c-b) \,\begin{vmatrix}b^2(c+a)&b&1\\a^2(c+b)&a&1\\a^2b^2&ab&a+b\end{vmatrix}\\ \stackrel{\large C_1\,-\,ab\,C_2}{\longrightarrow} &(c-a)(c-b)c \,\begin{vmatrix}b^2&b&1\\a^2&a&1\\0&ab&a+b\end{vmatrix}\\ \stackrel{\large R_1-R_2}{\longrightarrow} &(c-a)(c-b)c(b-a) \,\begin{vmatrix}b+a&1&0\\a^2&a&1\\0&ab&a+b\end{vmatrix}\\ \stackrel{\large R_2-aR_1}{\longrightarrow} &(c-a)(c-b)c(b-a) \,\begin{vmatrix}b+a&1&0\\-ab&0&1\\0&ab&a+b\end{vmatrix}=0. \end{align*}
Imagine expanding along the first column. Note that the cofactor of $b^2c^2$ is $$(a+b)ac-(a+c)ab=a^2(c-b)$$ which is a multiple of $a^2$. The other two terms in the expansion along the first column are certainly multiples of $a^2$, so the determinant is a multiple of $a^2$. By symmetry, it's also a multiple of $b^2$ and of $c^2$.
If $a=b$ then the first two rows are equal, so the determinant's zero, so the determinant is divisible by $a-b$. By symmetry, it's also divisible by $a-c$ and by $b-c$.
So, the determinant is divisible by $a^2b^2c^2(a-b)(a-c)(b-c)$, a poynomial of degree $9$. But the detrminant is a polynomial of degree $7$, so it must be identically zero.
