# Thread: Determinant with Very Hard Algebra

1. ## Determinant with Very Hard Algebra

The problem is
$\displaystyle \det \left [ \begin{matrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{matrix} \right ]$
Sorry I'm new. That's the best I could write the problem. It's easy to find the determinant with expansion by minors but to get the simplified solution involves some sort of clever trick that I don't know of. Here's what I can get to.
$\displaystyle abc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b)$
There are many ways to factor out terms from this.
$\displaystyle abc[bc(c - b) + ac(a - c) + ab(b - a)] =$
$\displaystyle abc[b(c + a)(c - a) + c(a + b)(a - b) + a(b + c)(b - c)] =$
$\displaystyle abc[c^2(b - a) + b^2(a - c) + a^2(c - b)]$
There are even more ways of course. None of these ways lead me to the simple solution:
$\displaystyle abc(b - a)(c - a)(c - b)$
Can someone explain how to factor this? Thanks.

2. Originally Posted by binarybob0001
The problem is
$\displaystyle \det \left [ \begin{matrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{matrix} \right ]$
Sorry I'm new. That's the best I could write the problem. It's easy to find the determinant with expansion by minors but to get the simplified solution involves some sort of clever trick that I don't know of. Here's what I can get to.
$\displaystyle abc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b)$
There are many ways to factor out terms from this.
$\displaystyle abc[bc(c - b) + ac(a - c) + ab(b - a)] =$
$\displaystyle abc[b(c + a)(c - a) + c(a + b)(a - b) + a(b + c)(b - c)] =$
$\displaystyle abc[c^2(b - a) + b^2(a - c) + a^2(c - b)]$
There are even more ways of course. None of these ways lead me to the simple solution:
$\displaystyle abc(b - a)(c - a)(c - b)$
Can someone explain how to factor this? Thanks.
Multiply out

\displaystyle \begin{aligned}(b - a)(c - a)(c - b) &= (bc -ac -ab +a^2)(c-b) \\ &= bc^2 - ac^2 -abc + a^2c -b^2c +abc + ab^2 - a^2b \\ &= bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b \end{aligned}

as you wanted.

3. You can only do that if you already know the solution. What if you didn't know the solution? I would have just stopped thinking I was done. Is there a systematic way of arriving at that result?

4. Hello, binarybob0001!

The problem is: .$\displaystyle \begin{bmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{bmatrix}$

Sorry I'm new. . . . . Welcome aboard!
That's the best I could write the problem. . . . . Good job!

It's easy to find the determinant with expansion by minors
but to get the simplified solution involves some sort of clever trick that I don't know of.

Here's what I can get to:
. . $\displaystyle abc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b)$

There are many ways to factor out terms from this.
None of these ways lead me to the simple solution:
. . $\displaystyle abc(b - a)(c - a)(c - b)$
Can someone explain how to factor this?
Here's an approach I "discovered" years ago . . .

Collect terms with respect to one of the variables.

Let's use $\displaystyle c$
. . Collect terms with $\displaystyle c^2$, and with $\displaystyle c$, and 'constants' (no $\displaystyle c$'s).

We have: .$\displaystyle abc\bigg[bc^2 - ac^2 - b^2c + a^2c + ab^2 - a^2b\bigg]$

Factor "by grouping": .$\displaystyle abc\bigg[c^2(b-a) - c(b^2-a^2) + ab(b-a)\bigg]$
. . $\displaystyle = \;abc\bigg[c^2(b-a) - c(b-a)(b+a) + ab(b-a)\bigg]$

Factor out $\displaystyle (b-a)\!:\;\;abc(b-a)\bigg[c^2 - c(b+a) + ab\bigg]$
. . $\displaystyle = \;abc(b-a)\bigg[c^2 - bc - ac + ab\bigg]$

Factor "by grouping": .$\displaystyle abc(b-a)\bigg[c(c-b) - a(c-b)\bigg]$

Factor out $\displaystyle (c-b)\!:\;\;abc(b-a)(c-b)(c-a)$ . . . . ta-DAA!

5. ahah, so I just need to chose a letter, stupid me. Thank you very much. Hey, are there more problems this hard to practice on?