MHB Divisibility of $(a+b+c)^{333}$ by $(a+b+c)^3$

[kaliprasad]

show that $(a+b+c)^{333}- a^{333} - b^{333} - c^{333}$ is divisible by $(a+b+c)^3 - a^3 - b^3 - c^3$

 

[anemone]

My solution:

Spoiler

If we simplify the denominator, we see that we have

$\begin{align*}(a+b+c)^3-a^3-b^3-c^3&=3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2abc)\\&=3(a+b)(b+c)(c+a)\end{align*}$

In fact, I "cheated" a bit because for the ugly expression such as the one above, noticed that one quick way to factor the expression is by letting $a=-b$, and most likely we would end up with zero after we replacing the $a$ by $-b$ or vice verse into the expression. Similarly, we see that we got the same result if we made the substitution of $b=-c$ and $c=-a$, and et voila, $(a+b)(b+c)(c+a)$ is the factored form of $(a+b+c)^3-a^3-b^3-c^3$.

But bear in mind that our goal is to prove that $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$ is divisible by $(a+b+c)^3-a^3-b^3-c^3$. Hence, by using the same trick I used above, notice that when

[TABLE="class: grid, width: 800"]
[TR]
[TD="width: 10%"]$a=-b$[/TD]
[TD]$\begin{align*}(a+b+c)^{333}-a^{333}-b^{333}-c^{333}&=(-b+b+c)^{333}+b^{333}-b^{333}-c^{333}\\&=c^{333}-c^{333}\\&=0\end{align*}$[/TD]
[/TR]
[TR]
[TD]$b=-c$[/TD]
[TD]$\begin{align*}(a+b+c)^{333}-a^{333}-b^{333}-c^{333}&=(a-c+c)^{333}-a^{333}+c^{333}-c^{333}\\&=a^{333}-a^{333}\\&=0\end{align*}$[/TD]
[/TR]
[TR]
[TD]$c=-a$[/TD]
[TD]$\begin{align*}(a+b+c)^{333}-a^{333}-b^{333}-c^{333}&=(a+b-a)^{333}-a^{333}-b^{333}+a^{333}\\&=b^{333}-b^{333}\\&=0\end{align*}$[/TD]
[/TR]
[/TABLE]

So we can say at this juncture that $(a+b)(b+c)(c+a)$ are three factors that come from factoring the numerator.

We are not finished yet, we still have to prove that the numerator, their value after the subtraction is divisible by 3. And this can be easily proved by the help of AM-GM formula.

Observe that $a+b+c\ge3\sqrt[3]{abc}\,\,\,\rightarrow(a+b+c)^{333}\ge3^{333}{abc}^{\dfrac{333}{3}}_{\phantom{i}}$ and $a^{333}+b^{333}+c^{333}\ge3(abc)^{\dfrac{333}{3}}_{\phantom{i}}$ (and for both inequalities, equality occurs when $a=b=c=1$) and this gives the minimum value of $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}\ge3^{333}{abc}^{\dfrac{333}{3}}_{\phantom{i}}-3(abc)^{\dfrac{333}{3}}_{\phantom{i}}\ge3{abc}^{\dfrac{333}{3}}_{\phantom{i}}(3^{332}-1)$

and now I think it's safe to conclude that $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$ is divisible by $(a+b+c)^3-a^3-b^3-c^3$.

 

[kaliprasad]

My solution:

Spoiler

If we simplify the denominator, we see that we have

$\begin{align*}(a+b+c)^3-a^3-b^3-c^3&=3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2abc)\\&=3(a+b)(b+c)(c+a)\end{align*}$

In fact, I "cheated" a bit because for the ugly expression such as the one above, noticed that one quick way to factor the expression is by letting $a=-b$, and most likely we would end up with zero after we replacing the $a$ by $-b$ or vice verse into the expression. Similarly, we see that we got the same result if we made the substitution of $b=-c$ and $c=-a$, and et voila, $(a+b)(b+c)(c+a)$ is the factored form of $(a+b+c)^3-a^3-b^3-c^3$.

But bear in mind that our goal is to prove that $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$ is divisible by $(a+b+c)^3-a^3-b^3-c^3$. Hence, by using the same trick I used above, notice that when

[TABLE="class: grid, width: 800"]
[TR]
[TD="width: 10%"]$a=-b$[/TD]
[TD]$\begin{align*}(a+b+c)^{333}-a^{333}-b^{333}-c^{333}&=(-b+b+c)^{333}+b^{333}-b^{333}-c^{333}\\&=c^{333}-c^{333}\\&=0\end{align*}$[/TD]
[/TR]
[TR]
[TD]$b=-c$[/TD]
[TD]$\begin{align*}(a+b+c)^{333}-a^{333}-b^{333}-c^{333}&=(a-c+c)^{333}-a^{333}+c^{333}-c^{333}\\&=a^{333}-a^{333}\\&=0\end{align*}$[/TD]
[/TR]
[TR]
[TD]$c=-a$[/TD]
[TD]$\begin{align*}(a+b+c)^{333}-a^{333}-b^{333}-c^{333}&=(a+b-a)^{333}-a^{333}-b^{333}+a^{333}\\&=b^{333}-b^{333}\\&=0\end{align*}$[/TD]
[/TR]
[/TABLE]

So we can say at this juncture that $(a+b)(b+c)(c+a)$ are three factors that come from factoring the numerator.

We are not finished yet, we still have to prove that the numerator, their value after the subtraction is divisible by 3. And this can be easily proved by the help of AM-GM formula.

Observe that $a+b+c\ge3\sqrt[3]{abc}\,\,\,\rightarrow(a+b+c)^{333}\ge3^{333}{abc}^{\dfrac{333}{3}}_{\phantom{i}}$ and $a^{333}+b^{333}+c^{333}\ge3(abc)^{\dfrac{333}{3}}_{\phantom{i}}$ (and for both inequalities, equality occurs when $a=b=c=1$) and this gives the minimum value of $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}\ge3^{333}{abc}^{\dfrac{333}{3}}_{\phantom{i}}-3(abc)^{\dfrac{333}{3}}_{\phantom{i}}\ge3{abc}^{\dfrac{333}{3}}_{\phantom{i}}(3^{332}-1)$

and now I think it's safe to conclude that $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$ is divisible by $(a+b+c)^3-a^3-b^3-c^3$.

solution looks good but

Spoiler

because it is algebra it is not required to prove that it is divisible by 3

and secondly your statement does not prove it it conclusively ( I might not have understood it)

 

[anemone]

I might not have submitted a great solution then, hehehe...and I'm looking forward to reading your solution in this case.

 

[kaliprasad]

I might not have submitted a great solution then, hehehe...and I'm looking forward to reading your solution in this case.

I am sorry about mis communication.

what I meant that

Spoiler

your solution 2nd part does not confirm that the expression is divisible by 3 . divisibility by (a+b)(b+c)(c+a) is confirmed.

 

[anemone]

I am sorry about mis communication.

what I meant that

Spoiler

your solution 2nd part does not confirm that the expression is divisible by 3 . divisibility by (a+b)(b+c)(c+a) is confirmed.

Now that I reread my solution, I realize what I previously said about the AM-GM helping to establish the fact that the numerator is divisible by 3 is complete nonsense which doesn't add up, and I am sorry, kali...

 

[kaliprasad]

Spoiler

first we factor $(a+b+c)^3 - a^3-b^3 - c^3$

$ (a+b+c)^3 - a^3 - b^3 - c^3$
=$a^3 + (b+c)^3 + 3a(b+c)(a+b+ c) - a^3 - b^3 - c^3$ using $(a+x)^3 = a^3 + x^3 + 3ax(a+x)$ x = b + c
= $a^3 + b^3 + c^3 + 3bc(b+c) + 3a(b+c)(a+b+ c) - a^3 - b^3 - c^3$
= $3bc(b + c)+ 3a(b+c)(a+b+ c)$
= $3(b+c)(bc + a^2 + ab + ac)$
= $3(b+c)(a+b)(c+a)$

now we need to show that $(a+b+c)^{333} - a^{333}-b^{333} - c^{333}$ is divisible by (a+b), (b+c) and (c+a)

$(a+b+c)^{333} - a^{333}-b^{333} - c^{333}$
= $((a+b+c)^{333} - a^{333})-(b^{333} + c^{333})$

as $ (a+b+c)^{333} - a^{333}) $ is divisible by (a+b+c) - a or b + c as $x^n-y^n$ is divisible by x-y

and $(b^{333} + c^{333})$ is divisible by b+ c as $x^n + y^n$ is dvisible by x+y for odd n

so $(a+b+c)^{333} - a^{333}-b^{333} - c^{333}$ is divisible by (b+c)

similarly it is divisible by a + b and c + a

as 3 is a constant we need not prove it as it is algebra

hence it is divisible by $(a+b+c)^3 - a^3-b^3 - c^3$