Re: Factorising this...

On Sun, 9 Dec 2007 17:09:13 -0800 (PST), Albert
<albert.xtheunknown0@xxxxxxxxx> wrote:

On Dec 10, 9:44 am, Rotwang <sg...@xxxxxxxxxxxxx> wrote:
On 9 Dec, 22:20, Albert <albert.xtheunkno...@xxxxxxxxx> wrote:



On Dec 10, 8:41 am, quasi <qu...@xxxxxxxx> wrote:

On Sun, 9 Dec 2007 13:30:21 -0800 (PST), Albert

<albert.xtheunkno...@xxxxxxxxx> wrote:
Hi, I'm in Year 8 (Australia) so this should be an easy question for
you guys seeing what this group is about... but how do you 'fully
factorise' this?

(They've given me the identities: difference of two squares, sum and
difference of two cubes and the perfect square rule but I'm not sure
how you apply them; i can do the individual questions relating to each
one of these indentities, but then again this question is in the
'problems section' [the last question] of this booklet...)

a²b² -a² -b² + 1

Hint:

First try 2 simpler problems:

(1) Factor: uv + u + v + 1

(2) Factor: uv - u - v + 1

quasi

Thanks - I solved my original question:

a²b² - b² -a² + 1 = b²(a² - 1) - (a² - 1)
= (a² - 1)(b² - 1)

Further factorization is possible - recall that x^2 - y^2 = (x + y)(x
- y).

Yep - [(a-1)(a+1)][(a-2)(a+2)]

?? Does the above make any sense? What happened to b?

Thanks

On a similar problem...

How would you factorise a ^ 6 - b ^ 6? Do you have to use the
'difference of two cubes' identity to solve it?

Yes, but hold off on that.

Instead, first look at it as a difference of 2 squares.

quasi
.

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