Use Remainder theorem to find factors of ##(a-b)^3+(b-c)^3+(c-a)^3##

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  • #1
chwala
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Homework Statement:
Use Remainder theorem to find factors of ##(a-b)^3+(b-c)^3+(c-a)^3##
Relevant Equations:
Remainder theorem
My first approach;
##(a-b)^3+(b-c)^3+(c-a)^3=a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^3+c^3-3c^2a+3ca^2-a^3##
##=-3a^2b+3ab^2-3b^2c+3bc^2-3c^2a+3ca^2##

what i did next was to add and subtract ##3abc## ...just by checking the terms ( I did not use Remainder theorem ):rolleyes:

##
=3abc-3a^2b+3ab^2-3b^2c+3bc^2-3c^2a+3ca^2-3abc##

##=3c(ab-ac-b^2+bc)-3a(ab-ac-b^2+bc)##
##=3(c-a)(ab-ac-b^2+bc)##
##=3(c-a)(a(b-c)-b(b-c))##
##=3(c-a)(a-b)(b-c)##

I need to check later on how to apply the Remainder theorem, any insight on this is welcome...

My second approach;

If##a=b## then ##f(b)=(b-c)^3+(c-a)^3##
If ##b=c##, ##f(c)=(a-b)^3+(c-a)^3##
If ##c=a##, ##f(a)=(a-b)^3+(b-c)^3##

If ##a=b## then ##(a-b)## will be a factor,
if ##b=c##, then ##(b-c) ## will be a factor,
If ##c=a##, then ##(c-a)## will be a factor. Therefore on taking product of the factors we shall have

##(a-b)(b-c)(c-a)=abc-a^2b-ac^2+a^2c-b^2c+ab^2+bc^2-abc## ... on multiplying both sides by ##3## we shall have,
##3(a-b)(b-c)(c-a)=3[abc-a^2b-ac^2+a^2c-b^2c+ab^2+bc^2-abc]≡(a-b)^3+(b-c)^3+(c-a)^3##
 
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Answers and Replies

  • #2
anuttarasammyak
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We observe it is quadratic for a, b and c and if any of two are equal it vanishes so it has factor (a-b)(b-c)(c-a).
say a=-1,b=0,c=1 it is 6 and (a-b)(b-c)(c-a)=2 so the coefficient is 3. In total 3 (a-b)(b-c)(c-a).
 
  • #3
fresh_42
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Homework Statement:: Use Remainder theorem to find factors of ##(a-b)^3+(b-c)^3+(c-a)^3##
Relevant Equations:: Remainder theorem

My approach;
##(a-b)^3+(b-c)^3+(c-a)^3=a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^3+c^3-3c^2a+3ca^2-a^3##
##=-3a^2b+3ab^2-3b^2c+3bc^2-3c^2a+3ca^2##

what i did next was to add and subtract ##3abc## ...just by checking the terms ( I did not use Remainder theorem ):rolleyes:

##
=3abc-3a^2b+3ab^2-3b^2c+3bc^2-3c^2a+3ca^2-3abc##

##=3c(ab-ac-b^2+bc)-3a(ab-ac-b^2+bc)##
##=3(c-a)(ab-ac-b^2+bc)##
##=3(c-a)(a(b-c)-b(b-c))##
##=3(c-a)(a-b)(b-c)##

I need to check later on how to apply the Remainder theorem, any insight on this is welcome...

Alternatively,
If##a=b## then ##f(b)=(b-c)^3+(c-a)^3##
If ##b=c##, ##f(c)=(a-b)^3+(c-a)^3##
If ##c=a##, ##f(a)=(a-b)^3+(b-c)^3##

If ##a=b## then ##(a-b)## will be a factor,
if ##b=c##, then ##(b-c) ## will be a factor,
If ##c=a##, then ##(c-a)## will be a factor. Therefore on taking product of the factors we shall have##(a-b)(b-c)(c-a)=abc-a^2b-ac^2+a^2c-b^2c+ab^2+bc^2-abc## ...

##f(b)-(c-a)^3=(b-c)^3##
##f(c)-(a-b)^3=(c-a)^3##
##f(a)-(b-c)^3=(a-b)^3##
What do you mean by remainder theorem? The Chinese remainder theorem, Euclidean division, or something else?

One immediately sees that the expression is zero whenever two out of ##a,b,c## coincide. Thus all ##a-b\, , \,b-c\, , \,c-a## divide the expression. They are also pairwise coprime, so ##(a-b)(b-c)(c-a)## divides the expression. Finally, consider the degrees.
 
  • #4
chwala
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What do you mean by remainder theorem? The Chinese remainder theorem, Euclidean division, or something else?

One immediately sees that the expression is zero whenever two out of ##a,b,c## coincide. Thus all ##a-b\, , \,b-c\, , \,c-a## divide the expression. They are also pairwise coprime, so ##(a-b)(b-c)(c-a)## divides the expression. Finally, consider the degrees.
https://www.purplemath.com/modules/remaindr.htm
 
  • #5
fresh_42
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Ok, that is Euclidean division. You can consider the polynomial consecutively as element of ##\mathbb{Z}[a]\, , \,\mathbb{Z}[ b ]\, , \,\mathbb{Z}[c] .## E.g. if ##p(a):=(a-b)^3+(b-c)^3+(c-a)^3\in \mathbb{Z}[a]## then we have (I change from ##a## to ##x## to make it clearer) ##p(x)=-3x^2b+3xb^2-b^3+(b-c)^3+3x^2c-3xc^2## which we want to divide by ##x-b## and ##c-x.##

I like your version better. At least we have to do it only once for symmetry reasons.

(Edit: Corrected the factor to ##x-b,x-c##.)
 
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