Mini-Challenge on a condition

Question

Hello I would like to purpose to you an enigma this is the following :

Let $a,b,c$ be positive real number find the condition on $abc$ and $ab+bc+ca$ and $a+b+c$ to have $a^9+b^9+c^9=3$

I have a solution using the following identity : $$a^9+b^9+c^9= 3a^3b^3c^3−45abc(ab+bc+ca)(a+b+c)^4+ 54abc(ab+bc+ca)^2(a+b+c)^2−27a^2b^2c^2(ab+bc+ca)(a+b+c) + (a+b+c)^9−9(ab+bc+ca)(a+b+c)^7+ 9(ab+bc+ca)^4(a+b+c)−30(ab+bc+ca)^3(a+b+c)^3+ 18a^2b^2c^2(a+b+c)^3+ 27(ab+bc+ca)^2(a+b+c)^5+ 9abc(a+b+c)^6−9abc(ab+bc+ca)^3$$

But I would like a solution without this because it's too ugly .

So could you help me ?

Answer 1

We can use the following identity to solve the problem in a more efficient manner than remembering a gigantic identity: $$a^3+b^3+c^3= \sigma_1^3-3\sigma_1\sigma_2+3\sigma_3$$ where $\sigma_1=a+b+c, \ \sigma_2=ab+bc+ca$, and $\sigma_3=abc$ are the Elementary Symmetric Polynomials. Next we find the cubed versions of the symmetric polynomials as follows $$a^3b^3+b^3c^3+c^3a^3=(ab+bc+ca)^3-3(ab+bc+ca)(a^2bc+b^2ac+c^2ab)+3a^2b^2c^2$$ $$a^3b^3+b^3c^3+c^3a^3=\sigma_2^3-3\sigma_1\sigma_3+3\sigma_3^2$$ and $$a^3b^3c^3=\sigma_3^3$$ Now we can plug these cubed versions back into our original identity to get $$a^9+b^9+c^9=(\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3)^3-3(\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3)(\sigma_2^3 - 3\sigma_1\sigma_3+3\sigma_3^2)+\sigma_3^9=3$$ You can then expand if you'd like to obtain your version, but I think I'll leave it here.

Answer 2

To make it less ugly, calculate the lower amounts instead, or setting new variables, for example:

${\begin{cases}a+b+c=x\\ab+bc+ca=y\\abc=z\\S_{n}=a^n+b^n+c^n\end{cases}}$

In fact, I have encountered this problem in a test that gives $x=3;y=-10;z=11$ and it asks me to find $S_{5}$. Here's how I attemped it (I have generalized it for $x;y;z$):

Step one:

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

$\Rightarrow a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=xy$

$\Rightarrow a^2(b+c)+b^2(c+a)+c^2(a+b)=xy-3z$

Step two:

$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=x^2-2y$

Step three:

$(a+b+c)(a^2+b^2+c^2)=x(x^2-2y)$

$\Rightarrow a^3+b^3+c^3+a^2(b+c)+b^2(c+a)+c^2(a+b)=x(x^2-2y)$

$\Rightarrow a^3+b^3+c^3+xy-3z=x^3-2xy$

$\Rightarrow a^3+b^3+c^3=x^3-3xy+3z$

Step four:

$a^4+b^4+c^4$

$=(a^2+b^2+c^2)^2-2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$

$=(a^2+b^2+c^2)^2-2((ab+bc+ca)^2-2a^{2}bc-2ab^{2}c-2abc^{2})$

$=(a^2+b^2+c^2)^2-2((ab+bc+ca)^2-2abc(a+b+c))$

$=(x^2-2y)^2-2(y^2-2zx)$

$=x^4-4x^{2}y+2y^2+4zx$

Step five:

$(a^3+b^3+c^3)(ab+bc+ca)=y(x^3-3xy+3z)$

$\Rightarrow a^4(b+c)+b^4(c+a)+c^4(a+b)+abc(a^2+b^2+c^2)=x^3y-3xy^2+3yz$

$\Rightarrow a^4(b+c)+b^4(c+a)+c^4(a+b)+z(x^2-2y)=x^3y-3xy^2+3yz$

$\Rightarrow a^4(b+c)+b^4(c+a)+c^4(a+b)+x^2z-2yz=x^3y-3xy^2+3yz$

$\Rightarrow a^4(b+c)+b^4(c+a)+c^4(a+b)=x^3y-x^2z-3xy^2+5yz$

Step six:

$(a+b+c)(a^4+b^4+c^4)=x(x^4-4x^{2}y+2y^2+4zx)$

$\Rightarrow a^5+b^5+c^5+a^4(b+c)+b^4(c+a)+c^4(a+b)=x^5-4x^3y+2xy^2+4x^2z$

$\Rightarrow a^5+b^5+c^5+x^3y-x^2z-3xy^2+5yz=x^5-4x^3y+2xy^2+4x^2z$

$\Rightarrow S_{5}=a^5+b^5+c^5=x^5-5x^3y+5xy^2+5x^2z-5yz$

You can continue similarly to $S_{9}$ to write the expression in terms of $x,y,z$, this will make the result much less ugly and also much easier to check. Thanks for taking time reading this long post.

Answer 3

If you set $e_1=a+b+c,$ $e_2=ab+ac+bc,$ and $e_3=abc$ and you define a recurrence relation as follows: \begin{eqnarray*} x_0 &=& 3 \\ x_1 &=& e_1 \\ x_2 &=& e_1^2-2e_2 \\ x_{n+3} &=& e_1x_{n+2} - e_2x_{n+1} + e_3x_n \end{eqnarray*} then it can be shown that $x_n = a^n+b^n+c^n.$ In order to have $a^9+b^9+c^9=3,$ there must be $x_0,\ldots,x_9$ such that $$\begin{pmatrix} 1 &&&&&&&&& \\ & 1 &&&&&&&& \\ && 1 &&&&&&& \\ -e_3 & e_2 & -e_1 & 1 &&&&&& \\ & -e_3 & e_2 & -e_1 & 1 &&&&& \\ && -e_3 & e_2 & -e_1 & 1 &&&& \\ &&& -e_3 & e_2 & -e_1 & 1 &&& \\ &&&& -e_3 & e_2 & -e_1 & 1 && \\ &&&&& -e_3 & e_2 & -e_1 & 1 & \\ &&&&&& -e_3 & e_2 & -e_1 & 1 \\ &&&&&&&&& 1\end{pmatrix} \begin{pmatrix} x_0 \\x_1 \\x_2 \\x_3 \\x_4 \\x_5 \\x_6 \\x_7 \\x_8 \\x_9 \end{pmatrix} = \begin{pmatrix} 3 \\ e_1 \\ e_1^2-2e_2 \\0\\0\\0\\0\\0\\0\\0\\3 \end{pmatrix} $$ In other words $$\det \begin{pmatrix} 1 &&&&&&&&&& 3\\ & 1 &&&&&&&&& e_1\\ && 1 &&&&&&&& e_1^2-2e_2\\ -e_3 & e_2 & -e_1 & 1 &&&&&&& \\ & -e_3 & e_2 & -e_1 & 1 &&&&&& \\ && -e_3 & e_2 & -e_1 & 1 &&&&& \\ &&& -e_3 & e_2 & -e_1 & 1 &&&& \\ &&&& -e_3 & e_2 & -e_1 & 1 &&& \\ &&&&& -e_3 & e_2 & -e_1 & 1 && \\ &&&&&& -e_3 & e_2 & -e_1 & 1 & \\ &&&&&&&&& 1 & 3 \end{pmatrix} = 0 $$