Maths question

[jimzee4]

Factorise a^2(b-c)+b^2(c-a)+c^2(a-b)
^2 means squared
Any ideas?

[factor-fiction]

Into what form? It has already been factorised in a way, but you can reorganis eit and factorise it in another way.
First expand all the brackets

= a²b - a²c+ b²c-b²a +c²a-c²b
= a²b -b²a +c²a - a²c+ b²c -c²b
= ab(a-b) +ac(c-a) + bc (b-c)

I'm not sure this is any better though than the original expression.

Are you sure you've stated the question exactly?

[Prudie]

If you mean factorise so it's all in brackets then you can do this (but I can't type out each line, take too long):
Multiply out first to get
a^2b - a^2c + b^2c - b^2a + c^2a - c^2b
You can then rearrange the order and get terms and take (b-c) as a factor (not doing it all for you) to give an answer of:

(b-c)(bc + a^2 - ac - ab)

[Prudie]

How do you do the squared symbol factor, found using that ^ very confusing?

[factor-fiction]

Yes I suppose you could take other combinations of Prudie's solution as well- eg take a-b as a factor.
However I think you can make 3 brackets.
I've not got time to work it out but if you juggle around with + and - signs you can try variations of (a-b)(a-c)(b-c). In fact that may be the answer

[factor-fiction]

Prudie- to get the ² sign I simply type it in Word then paste it into AB

[Woody23]

If you type it into Google, you'll get the answer.

[factor-fiction]

Can you show us that, woody?
Regards

[vascop]

^2(b-c)+b^2(c-a)+c^2(a-b)=a^2(b-c)+bc(b-c)-a(
b^2-c^2)

Since b^2-c^2=(b+c)(b-c) this becomes
a^2(b-c)+bc(b-c)-a(b+c)(b-c)
=(b-c){a^2+bc-a(b+c)}
=(b-c){a^2-ab+bc-ac}
Rearranging the stuff in the curly brackets to get a common factor of (a-c) gives
(b-c){a(a-c)-b(a-c)}
=(b-c)(a-c)(a-b)

[vascop]

Sorry that first line should be:
a^2(b-c)+b^2(c-a)+c^2(a-b)=a^2(b-c)+bc(b-c)-a
(b^2-c^2)

[jimzee4]

Thanks everybody. Only just saw the answers.
Too much footie.Will get back to the maths.