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# Factoring polynomials

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Factor \(ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a\).

I've gotten the factored version of the polynomial is -(a-b)(b-c)(c-a)(a+b+c), however I'm not really sure how to explain it. I know that we must begin with (a-b)(b-c)(c-a), but don't know where to go from there.

Thank you in advance.

Mar 14, 2021

### 10+0 Answers

#1
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This is easy.  Just plug it into Wolfram Alpha.

Mar 14, 2021
#2
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Yes, but I want to learn how to get to that answer :/

grs75  Mar 14, 2021
#3
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I'd like to see an answer to this one too. Mar 14, 2021
#4
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*Gasp* this one is not easy.

Beginning with (a-b)(b-c)(c-a) is correct.

That will get us (ab-ac-b^2+bc)(c-a) which is ab^2-a^2b-ac^2-b^2c+bc^2.

And then we can see that, ab^2-a^2b-ac^2-b^2c+bc^2 times -(a+b+c) will get us ab^3-a^3b+bc^3-b^3c+ca^3-c^3a.

So, we combine this to the factor we already got: (a-b)(b-c)(c-a) and get -(a+b+c)(a-b)(b-c)(c-a), which is the same as -(a-b)(b-c)(c-a)(a+b+c)

Mar 14, 2021
#5
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Hi Cal,

You have done some excellent expanding.  Good work.

But

I was not asking anyone to show that the answer was correct.

I was asking for someone to explain how the answer could be arrived at, if I did not already know it.

I'm pretty sure that is what grs75 wants too. Melody  Mar 15, 2021
#6
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Does this just depend on reasoning perhaps? I don't think there's anything complex, I simply got the answer by playing around but obviously that's not the correct way to solve it.

grs75  Mar 15, 2021
#7
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No Cal, you took the answer that was given and effectively checked that it was right.

You did not show how you could have got the answer for yourself.

You do not have enough maths knowledge to do that yet.  Maybe you will have one day Melody  Mar 15, 2021
#9
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i cRI  CalTheGreat  Mar 16, 2021
edited by CalTheGreat  Mar 22, 2021
#10
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Thanks for your input grs75.

If you did it yourself then you have more skill and knowledge than you are giving yourself credit for.

Melody  Mar 17, 2021
#8
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Oop yeah, I thought that grs75 already know how to get the (a-b)(b-c)(c-a) part.

In that case I'll gib some factoring tricks that can make this easier:

(a^b)^c = a^c*a^b

(a/b)^c=a^c/b^c

(a+b)^2=a^2+2ab+b^2

(a-b)^2=a^2-2ab+b^2

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

(a+b)^3=a^3+3a^2b+3ab^2+b^3

a^3-b^3=(a-b)(a^2+a+b^2)

a^3+b^3=(a+b)(a^2-ab+b^2)

I'm sure that if we rearrange this SOMEHOW we'll get it, but apparently I'm dumb today Mar 15, 2021