# Prove that is divisible by

Question:

Prove that is divisible by and find the quotient.

$\left|\begin{array}{lll}b c-a^{2} & c a-b^{2} & a b-c^{2} \\ c a-b^{2} & a b-c^{2} & b c-a^{2} \\ a b-c^{2} & b c-a^{2} & c a-b^{2}\end{array}\right|$

Solution:

$\Delta=\left|\begin{array}{ccc}b c-a^{2} & c a-b^{2} & a b-c^{2} \\ c a-b^{2} & a b-c^{2} & b c-a^{2} \\ a b-c^{2} & b c-a^{2} & c a-b^{2}\end{array}\right|$

Now, [Applying $C_{1} \rightarrow C_{1}-C_{2}$ and $\left.C_{2} \rightarrow C_{2}-C_{3}\right]$$\Delta=\left|\begin{array}{lll}b c-a^{2}-c a+b^{2} & c a-b^{2}-a b+c^{2} & a b-c^{2} \\ c a-b^{2}-a b+c^{2} & a b-c^{2}-b c+a^{2} & b c-a^{2} \\ a b-c^{2}-b c+a^{2} & b c-a^{2}-c a+b^{2} & c a-b^{2}\end{array}\right|$

$=\left|\begin{array}{lll}(b-a)(a+b+c) & (c-b)(a+b+c) & a b-c^{2} \\ (c-b)(a+b+c) & (a-c)(a+b+c) & b c-a^{2} \\ (a-c)(a+b+c) & (b-a)(a+b+c) & c a-b^{2}\end{array}\right|$

Next,

[Taking $(a+b+c)$ common from $C_{1}$ and $C_{2}$ each]

$\Delta=(a+b+c)^{2}\left|\begin{array}{ccc}b-a & c-b & a b-c^{2} \\ c-b & a-c & b c-a^{2} \\ a-c & b-a & c a-b^{2}\end{array}\right|$

Then,

[Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ ]

$\Delta=(a+b+c)^{2}\left|\begin{array}{ccc}0 & 0 & a b+b c+c a-\left(a^{2}+b^{2}+c^{2}\right) \\ c-b & a-c & b c-a^{2} \\ a-c & b-a & c a-b^{2}\end{array}\right|$

Lastly,

[Expanding along $R_{1}$ ]

$\Delta=(a+b+c)^{2}\left[a b+b c+c a-\left(a^{2}+b^{2}+c^{2}\right)\right]\left[(c-b)(b-a)-(a-c)^{2}\right]$

$=(a+b+c)^{2}\left(a b+b c+c a-a^{2}-b^{2}-c^{2}\right) \times$ $\left(b c-a c-b^{2}+a b-a^{2}-c^{2}+2 a c\right)$

$=(a+b+c)\left[(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)^{2}\right]$

Therefore, given determinant is divisible by $(a+b+c)$ and quotient is

$(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)^{2}$