1. ## Another factoring problem

Hello, it's me again with my factoring problems....

Problem 122. Factor:
(c) $\displaystyle a^3+b^3+c^3-3abc;$

My attempt: Only thing I was able to do is to expand $\displaystyle a^3+b^3=a^3+a^2b-a^2b+ab^2-ab^2+b^3$

Also, please explain the solution step by step.

2. ## Re: Another factoring problem

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

3. ## Re: Another factoring problem

Originally Posted by zaidalyafey
$\displaystyle a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$
Thanks for the reply, but I already found that on other sites, how you get that is my question.

4. ## Re: Another factoring problem

$\displaystyle a^3+b^3+c^3-3abc=a^3+b^3+c^3-abc-abc-abc$

add and subtract the terms $\displaystyle \,\,a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\,\,$ then use grouping

$\displaystyle a^3+a^2b+a^2c+b^3+b^2a+b^2c+c^3+c^2a+c^2b-abc-a^2b-ab^2-abc-c^2b-b^2c-abc-a^2c-ac^2$

$\displaystyle a^2(a+b+c)+b^2(a+b+c)+c^2(a+b+c)-ab(a+b+c)-bc(a+b+c)-ac(a+b+c)$

$\displaystyle (a+b+c)(a^2+b^2+c^2-ab-ac-bc)$

5. ## Re: Another factoring problem

a^3+b^3=(a+b)(a^2-ab+b^2 )
Sub b=b+c
a^3+(b+c)^3=(a+b+c)(a^2-a(b+c)+(b+c)^2 )
a^3+b^3+〖3b〗^2 c+〖3bc〗^2+c^3=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )
a^3+b^3+c^3=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-〖3b〗^2 c-〖3bc〗^2
a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-〖3b〗^2 c-〖3bc〗^2-3abc
a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-3bc(a+b+c)
a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2-3bc)
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)

This is one method of getting that hope you understand...

6. ## Re: Another factoring problem

Originally Posted by zaidalyafey
$\displaystyle a^3+b^3+c^3-3abc=a^3+b^3+c^3-abc-abc-abc$

add and subtract the terms $\displaystyle \,\,a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\,\,$ then use grouping

$\displaystyle a^3+a^2b+a^2c+b^3+b^2a+b^2c+c^3+c^2a+c^2b-abc-a^2b-ab^2-abc-c^2b-b^2c-abc-a^2c-ac^2$

$\displaystyle a^2(a+b+c)+b^2(a+b+c)+c^2(a+b+c)-ab(a+b+c)-bc(a+b+c)-ac(a+b+c)$

$\displaystyle (a+b+c)(a^2+b^2+c^2-ab-ac-bc)$
Thank you for the explanation! That was really helpful!

7. ## Re: Another factoring problem

on thing to notice about $\displaystyle a^3 + b^3 + c^3 - 3abc$ is that it is symmetric in a,b and c: switch any two of a,b, and c and you still get the same thing.

so if it factors, it will factor as two symmetric expressions, one involving just a single term at a time, and one involving two terms at a time.

but the only symmetric expression in a,b,c involving just a single term at a time is: a+b+c, or some multiple thereof. so let's divide $\displaystyle a^3 + b^3 + c^3 - 3abc$ by a+b+c,

considering it as a polynomial in a.

obviously the first term of the quotient is going to be $\displaystyle a^2$, so we get:

$\displaystyle (a + b + c)(a^2) = a^3 + a^2b + a^2c$. subtracting that from $\displaystyle a^3 + b^3 + c^3 - 3abc$ gives us:

$\displaystyle -(b+c)a^2 - (3bc)a + b^3 + c^3$. the $\displaystyle a^2$ term's coefficient is $\displaystyle -(b+c)$, so we take:

$\displaystyle (a+b+c)(-(b+c)a) = -(b+c)a^2 - (b^2 + bc)a - (bc + c^2)a$

$\displaystyle = -(b+c)a^2 - (b^2 + c^2 + 2bc)a$, and subtract THAT from our previous remainder:

$\displaystyle -(b+c)a^2 - (3bc)a + b^3 + c^3 -(-(b+c)a^2 - (b^2 + c^2 + 2bc)a)$

$\displaystyle = -(b+c)a^2 - (3bc)a + b^3 + c^3 + (b+c)a^2 + (b^2 + c^2 + 2bc)a$

$\displaystyle = (b^2 + c^2 - bc)a + b^3 + c^3$.

now the coefficient of a is $\displaystyle b^2 + c^2 - bc$, so we take:

$\displaystyle (a+b+c)(b^2 + c^2 - bc) = (b^2 + c^2 - bc)a + b^3 + bc^2 - b^2c + b^2c + c^3 - bc^2$

$\displaystyle = (b^2 + c^2 - bc)a + b^3 + c^3$.

as this is equal to our last remainder, we have:

$\displaystyle a^3 + b^3 + c^3 - 3ab = (a+b+c)(a^2 -(b+c)a + b^2 + c^2 - bc)$

or, re-arranging a little:

$\displaystyle a^3 + b^3 + c^3 - 3ab = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc)$.

8. ## Re: Another factoring problem

Another approach:

$\displaystyle a^3 + b^3 + c^3 -3abc = \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix} = \begin{vmatrix}a+b+c & b & c \\ a+b+c & a & b \\ a+b+c & c & a \end{vmatrix} = (a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & a & b \\ 1 & c & a \end{vmatrix} = (a+b+c) (a^2 + b^2 + c^2 - ab - bc - ac)$

Source: "My Favourite Polynomial", Desmond MacHale, The Mathematical Gazette, Vol. 75, No. 472 (June 1991), pp. 157-165.

9. ## Re: Another factoring problem

Originally Posted by sujith123
a^3+b^3=(a+b)(a^2-ab+b^2 )
Sub b=b+c
a^3+(b+c)^3=(a+b+c)(a^2-a(b+c)+(b+c)^2 )
a^3+b^3+〖3b〗^2 c+〖3bc〗^2+c^3=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )
a^3+b^3+c^3=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-〖3b〗^2 c-〖3bc〗^2
a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-〖3b〗^2 c-〖3bc〗^2-3abc
a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-3bc(a+b+c)
a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2-3bc)
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)

This is one method of getting that hope you understand...
Thank you, your explanation was the easiest to understand.