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# Polynomial help

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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.

Jul 14, 2024

#1
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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:

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

Jul 14, 2024
#3
+1790
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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! :)

Jul 22, 2024
edited by NotThatSmart  Jul 22, 2024