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

[chwala]

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 )

$=3abc-3a^2b+3ab^2-3b^2c+3bc^2-3c^2a+3ca^2-3abc$
$=3c(ab-ac-b^2+bc)-3a(ab-ac-b^2+bc)$
$=3(c-a)(ab-ac-b^2+bc)$
$=3(c-a)(a(b-c)-b(b-c))$
$=3(c-a)(a-b)(b-c)$

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,
$3(a-b)(b-c)(c-a)=3[abc-a^2b-ac^2+a^2c-b^2c+ab^2+bc^2-abc]≡(a-b)^3+(b-c)^3+(c-a)^3$

 

[anuttarasammyak]

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).

 

[fresh_42]

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 )

$=3abc-3a^2b+3ab^2-3b^2c+3bc^2-3c^2a+3ca^2-3abc$
$=3c(ab-ac-b^2+bc)-3a(ab-ac-b^2+bc)$
$=3(c-a)(ab-ac-b^2+bc)$
$=3(c-a)(a(b-c)-b(b-c))$
$=3(c-a)(a-b)(b-c)$

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.

 

[chwala]

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

 

[fresh_42]

https://www.purplemath.com/modules/remaindr.htm

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 $x-b,x-c$.)