Show that $a+b+c$ divides at least one of the following numbers: $a^3+b^3-2c^3, b^3+c^3-2a^3, a^3+c^ 3-2b^3$

Question

The question

The natural numbers $a,b,c$ are considered non-zero and distinct two by two, so that $a^3+b^3+c^3=\frac{a^2b^2}{c}+\frac{a ^2c^2}{b}+\frac{b^2c^2}{a}$

a) It is possible that one of the numbers is equal to the gemometric average of the other two?

b) Show that $a+b+c$ divides at least one of the following numbers: $a^3+b^3-2c^3, b^3+c^3-2a^3, a^3+c^ 3-2b^3$

my idea

I was able to do the first point.

a) Let's verify if $a^3+b^3+c^3=\frac{a^2b^2}{c}+\frac{a ^2c^2}{b}+\frac{b^2c^2}{a}$, knowing that $a=\sqrt{bc}$

The equality would look like

$(\sqrt{bc})^3+b^3+c^3=\frac{b^3*c}{c}+\frac{c^3*b}{b}+\frac{b^2c^2}{\sqrt{bc}}$

$(\sqrt{bc})^3+b^3+c^3=b^3+c^3+\frac{b^2c^2\sqrt{bc}}{bc}$

$(\sqrt{bc})^3=bc\sqrt{bc}$ which is totally true, so this means that it is possible that one of the numbers is equal to the gemometric average of the other two.

b) For point b I thought writing the equality as a product where we see the numbers $a^3+b^3-2c^3, b^3+c^3-2a^3, a^3+c^ 3-2b^3$ and on the other side of it $a+b+c$ but I got only on wrong or unuseful paths,

Hope one of you can help me! Thank you!

Answer 1

Multiply both parts of $a^3+b^3+c^3=\frac{a^2b^2}{c}+\frac{a ^2c^2}{b}+\frac{b^2c^2}{a}$ by $abc$:

$$a^4bc+ab^4c+abc^4-a^3b^3-b^3c^3-c^3a^3=0.$$

Now expand the following: $$(a^2-bc)(b^2-ca)(c^2-ab)=$$

$$= a^4bc+ab^4c+abc^4-a^3b^3-b^3c^3-c^3a^3=0.$$

So without loss of generality let us assume that $a^2=bc$ (the part (a) was a hint all along). Then $$c^3+b^3-2a^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc-3a^3=$$ $$= (a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$

We are done.

Answer 2

I am almost completely sure that there exists a way better solution than my very tedious and boring solution but this is the only one that comes to mind so sorry about that :

firstly start off by mutliplying the denominators in the given relation to get :

$abc \cdot (a^3 + b^3 + c^3) = (ab)^3 + (bc)^3 + (ac)^3$ and subtract $3a^2b^2c^2$on both sides (and use $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2)$) to get :

$abc(a+b+c)(a^2+b^2+c^2-ab-bc-ac) = (ab+bc+ac)((ab)^2+(bc)^2+(ac)^2-ab^2c-ac^2b-ca^2b)$ and then take abc common from RHS to get :

$abc(a+b+c)(a^2+b^2+c^2-ab-bc-ac) = abc(ab+bc+ac)(\frac{ab}{c}+\frac{cb}{a}+\frac{ac}{b}-(a+b+c))$

cancel abc from both sides and open the brackets partially to get :

$(a+b+c)(a^2+b^2+c^2) - (a+b+c)(ab+bc+ac) = (ab+bc+ac)(\frac{ab}{c}+\frac{cb}{a}+\frac{ac}{b}) -(ab+bc+ac)(a+b+c)$

cancel (a+b+c)(ab+bc+ac) on both sides to finally get :

$(a+b+c)(a^2+b^2+c^2) = (ab+bc+ac)(\frac{ab}{c}+\frac{cb}{a}+\frac{ac}{b})$

add 2(a+b+c)(ab+bc+ac) on both sides to get :

$(a+b+c)(a^2+b^2+c^2 + 2(ab+bc+ab)) = (ab+bc+ac)(\frac{ab}{c}+\frac{cb}{a}+\frac{ac}{b}+2(a+b+c))$

$\implies (a+b+c)^3 = (ab+bc+ac)(\sqrt\frac{ab}{c}+\sqrt\frac{cb}{a}+\sqrt\frac{ac}{b})^2$

take lcm on RHS and simplify to get :

$(a+b+c)^3 = \frac{(ab+bc+ac)^3}{abc}$

$\implies a+b+c = \frac{ab + bc + ac}{{(abc)}^{\frac{1}{3}}}$

Now let $ac = k^3b^2$ (ofcourse k = 1 satisfies the queation as you proved in your post)

$\implies a+b+c = \frac{ab + bc + ac}{kb} = \frac{a+c}{k} + k^2b$ (thus k is also a natural number)

$\implies (a + c)(1-\frac{1}{k}) = b(k^2-1)$

again taking an lcm and using $(k^2-1) = (k-1)(k+1)$, we get :

$\frac{(a+c)(k-1)}{k} = b(k-1)(k+1)$

$\implies a+c = k(k+1)b$

$\therefore a+b+c = b(k^2+k+1)$

let us prove $a^3+c^3-2b^3 | a+b+c$, so we need :

$a^3+c^3-2b^3 = (a+b+c)(\text{some natural number})$

Use $a+b+c = b(k^2+k+1)$ and divide by b to get :

$\frac{a^3+c^3}{b}-2b^2 = (k^2+k+1)(\text{some natural number})$

$a^3+c^3 = (a+c)(a^2+c^2-ac) = (a+c)((a+c)^2-3ac)$

Using $(a+c) = (k^2+k)b \;\&\; ac =k^3b^2$ we get :

$\frac({a+c}{b} = k^2+k$

$\implies \frac{a^3+c^3}{b}-2k^3b^2 = (k^2+k)(b^2(k^2+k)^2-3k^3b^2) = (k^2+k+1)(\text{some natural number})$

$\therefore \frac{b^2((k^2+k)^3-3k^4(k+1)-2)}{k^2+k+1} = \text{some natural number}$

Open the inner brackets and simplify to get $k^6 + k^3 - 2 = (k^3+2)(k^3-1) = (k^3-2)(k-1)(k^2+k+1)$

$\therefore \frac{b^2((k^2+k)^3-3k^4(k+1)-2)}{k^2+k+1} = \frac{b^2(k^3+2)(k-1)(k^2+k+1)}{k^2+k+1} = b^2(k^3+2)(k-1) = \text{some natural number}$

Since b,k are natural numbers, $b^2, k^3+2 \;\&\; k-1$ are also natural numbers (if k = 1 then k-1 = 0 which also holds for the given condition)

Hence proved $b^2(k^3+2)(k-1) = \text{some natural number}$ and since this is true, our original assumption is also true that $a+b+c \; \text{divides} \; a^3 + c^3 - 2b^3$