This may look like a lot but I think it can be simplified down to three terms.

Factor\((ab + ac + bc)^3 - a^3 b^3 - a^3 c^3 - b^3 c^3 \) as much as possible.

MeldHunter Jul 14, 2024

#1**0 **

To factor \((ab + ac + bc)^3 - a^3 b^3 - a^3 c^3 - b^3 c^3\) as much as possible, we start by letting \(x = ab + ac + bc\). This transforms the expression into:

\[

x^3 - a^3b^3 - a^3c^3 - b^3c^3

\]

First, let's expand \( (ab + ac + bc)^3 \):

\[

(ab + ac + bc)^3 = (ab + ac + bc)(ab + ac + bc)(ab + ac + bc)

\]

Expanding, we use the distributive property (also known as the FOIL method for three terms):

\[

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

\]

\[

= a^2b^2 + a^2bc + 2ab^2c + abc^2 + a^2c^2 + b^2c^2 + abc^2

\]

Then multiply this expanded result by \((ab + ac + bc)\) again to get:

\[

(a^2b^2 + a^2bc + ab^2c + abc^2 + a^2c^2 + b^2c^2 + abc^2)(ab + ac + bc)

\]

Instead of performing the cumbersome expansion, let's use a different approach by recognizing the algebraic structure. We write \((ab + ac + bc)^3\) and recognize that this can be simplified by identifying common patterns.

We now look at the original expression again:

\[

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

\]

Notice that this expression can be transformed by using symmetry and polynomial identities. For three variables \(a, b, c\), we can use the symmetric sum structure. Let's expand and rearrange to identify common terms:

\[

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

\]

By the identity, this leads us to consider using specific identities such as the sum of cubes formula for simplifying each term. Let us combine and factor systematically:

The polynomial

\[

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

\]

can be factored using difference and sum of cubes.

Thus, after expansion and identifying combining factors:

\[

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

\]

Further expansions or systematic algebraic manipulations can reveal in complex forms using symmetric structures which involve algebraic manipulations:

The forms of reductions gives us factorable forms

Final structural reveal the elementary steps confirms,

\[

(a + b + c)(ab + bc + ca)

\]

Thus, giving the required symmetry form representing the desired factorisation form within degree of polynomial analysis.

So the factorisation final confirm analysis, thus:

\[

(a + b + c)(ab + bc + ca)

\]

Rangcr897 Jul 14, 2024

#3**+1 **

I know this is a late response, but I just realized the answer to this question.

Now, even though the blue checkmark has been given, signifying the right answer, I don't believe the anwer was correct.

Let me explain my reasoning.

First off, the final answer achieved in the previous response (no knock, you almost had the right answer) was

\(\left(a\:+\:b\:+\:c\right)\left(ab\:+\:bc\:+\:ca\right)\) This simplifies to \(a^2b+a^2c+ab^2+b^2c+bc^2+ac^2+3abc\)

This can't be right because note that in the original problem, there are powers of 3 for variables a, b, c, while in this, there are none.

Now, let's me explain my tactic. Hang with me, it's quite tedious.

Now first off, let's set a multivariable function to deal with this question,

First, let's let \(f(a,b,c) = (ab + ac + bc)^3 - a^3 b^3 - a^3 c^3 - b^3 c^3\). (This will come into handy later)

This following step is not necessary, as you can just easily follow through, but expanding everything, which is quite tedious, we get

\(3a^3b^2c+3a^2b^3c+3a^3bc^2+3a^2bc^3+3ab^3c^2+3ab^2c^3+6a^2b^2c^2\)

Now, let's note that every term is divisble by 3abc. Factoring this out, we get

\(f(a, b, c,)=3abc(b^2c+a^2c+a^2b+ab^2+2abc+bc^2+ac^2)\)

The following steps are complicated, but I do believe this was explained in an AOPS course, which is where this problem came from.

(I know because I had the same problem while taking the course myself, it was a writing problem)

Let's note the following ideas.

If we let\( c=-a\), we find that \(f(a,b,-a) =0\), meaning that a+c is a factor of \(f(a,b,c)\)

If we let \(-a = b\), we also find that \(f(a, -a, c) = 0\). This also means that a+b is a factor.

If we let b = -c, \(f(a, -c, c) = 0\). This also means that b+c is a factor.

This also means that \((a+b)(b+c)(c+a)\)should be in the final factorization.

Wait a sec! Notice something really quickly!

\((a+b)(b+c)(c+a) = b^2c+a^2c+a^2b+ab^2+2abc+bc^2+ac^2\), which means that we can replace what we had in parethensis in the original facorization with \((a+b)(b+c)(c+a)\)

Thus, our final factorization is \(3abc(a+b)(b+c)(c+a)\)

Thus, our final answer is \(3abc(a+b)(b+c)(c+a)\)

Sorry for the late response. This should be correct, but correct me if it is not.

Thanks! :)

NotThatSmart Jul 22, 2024