# Proving determinants using properties of determinants

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

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

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It's just a computation, have you made any progress? – Najib Idrissi Aug 30 '14 at 10:26
no actuall we have to show the both are equal without expansion – Raza Hassan Aug 30 '14 at 11:04
Subtract the first row from the second and the third row. This produces two zeros and you only have to calculate a $2\ \times \ 2$ - determinant. – Peter Aug 30 '14 at 11:13
That is the kind of information you are supposed to put in your question in the first place; all important information belongs in your question. Next, can you show any thoughts, ideas, work or attempts? – blue Aug 30 '14 at 11:14

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.

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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.

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but according to the question the factoe should be (a-b) and not (b-a) – Raza Hassan Aug 30 '14 at 13:36
@RazaHassan: $(b-a)=-(a-b)$, you can take care of the sign later. It happened that factoring out $b-a$ gave a more pleasant result, so I preferred this to reduce the chance of making an error doing the mental computation. – Marc van Leeuwen Aug 30 '14 at 13:57

@Raza 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 factorises.

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how exactly i don't get you – Raza Hassan Aug 30 '14 at 13:38
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – amWhy Aug 30 '14 at 17:28
@amWhy Simply giving the detailed answer to a straightforward homework-type question is not always the best way to help someone. After a few hours Raza put up his attempt. I pinpointed the error. If he fixed that, it would be straightforward to finish it. – almagest Aug 30 '14 at 18:21