Proving determinants using properties of determinants

Question

$$\begin{vmatrix} 1 & a^2+bc & a^3\\ 1 & b^2+ca & b^3\\ 1 & c^2+ab & c^3 \end{vmatrix} = (a-b)(b-c)(c-a)(a^2+b^2+c^2)$$

we have to solve this by using the properties of determinants without actually expanding the determinant. I am Unable to think which calculation to apply so was hoping for an hint.

Edit: just tried the problem and here is how I have done it

$$\begin{vmatrix} 1 & a^2+bc & a^3\\ 1 & -(a^2+b^2)+c(a-b) & -(a^3-b^3)\\ 1 & c^2-a^2-b(c-a) & c^3-a^3 \end{vmatrix}$$

then

$$\begin{vmatrix} 1 & a^2+bc & a^3\\ 0 & -((a-b)(a+b))+c(a-b) & -((a-b)(a^2+ab+b^2))\\ 0 & (c-a)(c+a)-b(c-a) & (c-a)(c^2+ca+a^2) \end{vmatrix}$$

then

$$(a-b)(c-a)\begin{vmatrix} 1 & a^2+bc & a^3\\ 0 & a+b+c & -(a^2+ab+b^2)\\ 0 & a-b+c & (c^2+ca+a^2) \end{vmatrix}$$

then $$(a-b)(c-a)\begin{vmatrix} a+b+c & -(a^2+ab+b^2)\\ a-b+c & (c^2+ca+a^2) \end{vmatrix}$$

then

$$(a-b)(c-a) * ( (a+b+c)(c^2+ca+a^2)-(a-b+c)(-(a^2+ab+b^2)) )$$

Here I am Confused on how to multiply them and get the answer

note: typed because I own a very bad handwriting

Thank you every one for your help

Answer 1

The first trick is to get as much zeroes as you can in the first row. That makes multiplication easier.

$\begin{vmatrix} 1 & a^2+bc & a^3 \\ 1 & b^2+ca & b^3\\ 1 & c^2+ab &c^3 \end{vmatrix}$

subtracting second row from first row and third row from second row:

$(a-b)(b-c)\begin{vmatrix} 0 & a+b-c & a^2+ab+b^2 \\ 0 & b+c-a & b^2+bc+c^2\\ 1 & c^2+ab &c^3 \end{vmatrix}$

Subtracting second row from first row:

$-(a-b)(b-c)(c-a)\begin{vmatrix} 0 & 2 & a+b+c \\ 0 & b+c-a & b^2+bc+c^2\\ 1 & c^2+ab &c^3 \end{vmatrix}$

exchanging row and column:

$(a-b)(b-c)(c-a)\begin{vmatrix} 0 & 0 & 1 \\ 2 & b+c-a & c^2+ab\\ a+b+c & b^2+bc+c^2 &c^3 \end{vmatrix}$

Now take determinant and get the result.

Answer 2

Seeing the column of $1$'s, a first thought would be to subtract the first row from the other two. This turns out to give you a factor $b-a$ in the second row and a factor $c-a$ in the third. $$ \begin{align} \begin{vmatrix} 1 & a^2+bc & a^3\\ 1 & b^2+ca & b^3\\ 1 & c^2+ab & c^3 \end{vmatrix} &= \begin{vmatrix} 1 & a^2+bc & a^3\\ 0 & b^2-a^2+c(a-b) & b^3-a^3\\ 0 & c^2-a^2+b(a-c) & c^3-a^3 \end{vmatrix} \\{}& =(b-a)(c-a) \begin{vmatrix} 1 & a^2+bc & a^3\\ 0 & (b+a)-c & a^2+ab+b^2\\ 0 & (c+a)-b & a^2+ac+c^2 \end{vmatrix} \end{align} $$ Now you can subtract the second row from the third, which makes the latter divisible by $b-c$, so you can factor again. I shouldn't be too difficult to complete the computation.

Answer 3

The solution-attempt in the question is excellent (after your edit)!

But unfortunately you made a small mistake. After taking out the factor $(a-b)(c-a)$, the middle element should be $c-a-b$, not $a+b+c$. If you fix that, the remaining determinant can be expanded and easily factorised.