- #1

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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 )

##

=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##

##(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##

Last edited: