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

In summary, the conversation discusses different approaches to finding the factors of the polynomial (a-b)^3+(b-c)^3+(c-a)^3, including using the Remainder theorem and considering the expression as elements of different polynomial rings. It is determined that the factors are (a-b)(b-c)(c-a), and the degrees are considered to support this conclusion.
  • #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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  • #2
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
chwala said:
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
fresh_42 said:
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
chwala said:
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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What is the Remainder theorem?

The Remainder theorem states that if a polynomial is divided by a linear polynomial, the remainder is equal to the value of the polynomial at the root of the linear polynomial.

How do we use the Remainder theorem to find factors of a polynomial?

To use the Remainder theorem to find factors of a polynomial, we first need to identify the linear polynomial that will serve as the divisor. Then, we divide the given polynomial by the linear polynomial and set the remainder equal to zero. The resulting equation can be solved to find the roots, which will be the factors of the polynomial.

What is the formula for the Remainder theorem?

The formula for the Remainder theorem is: If a polynomial f(x) is divided by (x-a), then the remainder is f(a).

Can the Remainder theorem be applied to any polynomial?

Yes, the Remainder theorem can be applied to any polynomial as long as the divisor is a linear polynomial.

How can we use the Remainder theorem to find factors of ##(a-b)^3+(b-c)^3+(c-a)^3##?

To find factors of ##(a-b)^3+(b-c)^3+(c-a)^3## using the Remainder theorem, we first need to identify the linear polynomial that will serve as the divisor. In this case, the linear polynomial will be (a-b)+(b-c)+(c-a) = 0. Then, we divide the given polynomial by (a-b)+(b-c)+(c-a) = 0 and set the remainder equal to zero. The resulting equation can be solved to find the roots, which will be the factors of the polynomial.

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