$ \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) $
I have a question. I tried so to solve it, but there is a problem.
that is i don't have any idea to findout how can i work with degrees 4,5,7 ...
this is the problem :
let $ x , y $ and $ z $ three real numbers such $ x+y+z = 0 $.
prove : $ \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) $
Please think and write your solutions! ; )
HINT:
Let $x,y,z$ be the roots of $\displaystyle t^3+bt+c=0\ \ \ \ (1)$
$\displaystyle\implies xy+yz+zx=b, xyz=-c$
Multiplying $(1)$ by $t^n\ne0$
$\displaystyle \implies t^{n+3}+bt^{n+1}+ct^n=0$
$\displaystyle \implies\sum x^{n+3}=-b\sum x^{n+1}-c\sum x^n$
$\displaystyle n=1\implies \sum x^4=-b\sum x^2-c\sum x =-b[(\sum x)^2-2(xy+yz+zx)]=-b(-2b)=2b^2$
From $\displaystyle n=0\implies \sum x^3=-b\sum x-3c=-3c$
$\displaystyle n=2\implies \sum x^5=-b\sum x^3-c\sum x^2 =-b(-3c)-c[(\sum x)^2-2(xy+yz+zx)]=3bc-c(-2b)=5bc$
Can you take home from here?
If $x=y=z=0$, then it is trivial.
WLOG, $x\ne 0$ (at least one of variables $\ne 0$).
Denote
$y=ax$,
$z=-x(1+a)$.
It is enough to prove, that
$$ \left(\frac{1^7+a^7-(1+a)^7}{7}\right)^2 = \left( \frac{1^5+a^5-(1+a)^5}{5} \right)^2 \cdot \left(\frac{1^4+a^4+(1+a)^4}{2}\right); $$
$$ \left(\frac{7a+21a^2+35a^3+35a^4+21a^5+7a^6}{7}\right)^2 = \left( \frac{5a+10a^2+10a^3+5a^4}{5} \right)^2 \cdot \left(\frac{2+4a^2+6a^3+4a^4+2a^5}{2}\right); $$
$$ (1+3a+5a^2+5a^3+3a^4+a^5)^2 = (1+2a+2a^2+a^3)^2 \cdot (1+2a+3a^2+2a^3+a^4); $$
$$ (1+3a+5a^2+5a^3+3a^4+a^5) = (1+2a+2a^2+a^3) \cdot (1+a+a^2). $$
Last identity is obvious.
Let $x,y,z$ be the roots of $t^3-Qt - P=0$, $Q= -(xy+yz+zx), P = xyz$.
$0 = (x+y+z)^2 = -2Q + x^2+y^2+z^2 \implies x^2+y^2+z^2 = 2Q$.
You have $t^3 = P + Qt$ and when you replace with $x,y,z$ and sum: $$ x^3+y^3+z^3 = 3P + Q(x+y+z) = 3P $$ You also get the identities: $$ x^4+y^4+z^4 = Q(x^2+y^2+z^2) = 2Q^2 \\ x^{n+3}+y^{n+3}+z^{n+3} = P(x^n+y^n+z^n) + Q(x^{n+1}+y^{n+1}+z^{n+1}) $$
Now return to the problem. Define $S_n = x^n+y^n+z^n$.
$$ LHS = \frac 1{49}S_7^2 = \frac 1{49}(PS_4+ QS_5)^2 = \frac 1{49}(2PQ^2 + Q(PS_2 + QS_3))^2\\ = \frac 1{49}(2PQ^2 + 2PQ^2 + 3PQ^2 )^2 = P^2Q^4 $$
$$ RHS = \frac1{50}S_5^2S_4 = \frac1{50}(PS_2+QS_3)^2 \times 2Q^2\\ = \frac1{50}(2PQ +3PQ)^2 \times 2Q^2 = P^2Q^4 = LHS $$
Given that $x + y + z = 0 $ , so, we have : $$\left(x+y+z \right)^2 = x^2 + y^2 + z^2 + 2 (xy + yz + zx) $$
So, we get : $$x^2 + y^2 + z^2 = \left(x+y+z\right)^2 - 2(xy + yz + zx) = -2(xy + yz + zx) $$ Squaring both sides,we get : $$\left(x^2 + y^2 + z^2 \right)^2 = 4(x^2 y^2 + y^2 z^2 + z^2 x^2 + 2(xy^2 z + yz^2 x + zx^2 y)) $$ Again, simplifying : $$ \begin {align} & x^4 + y^4 + z^4 + 2 (x^2 y^2 + y^2 z^2 + z^2 x^2 ) = 4(x^2 y^2 + y^2 z^2 + z^2 x^2 + 2(xy^2 z + yz^2 x + zx^2 y)) \\ & x^4 + y^4 + z^4 = 2(x^2 y^2 + y^2 z^2 + z^2 x^2) + 8(xy^2 z + yz^2 x + zx^2 y) \end{align} $$
Now, we can also write this as : $$ \begin{align} & \cfrac{x^4 + y^4 + z^4}{2} = x^2 y^2 + y^2 z^2 + z^2 x^2 + 4(xy^2 z + yz^2 x + zx^2 y) \\ & \cfrac{x^4 + y^4 + z^4}{2} = x^2 y^2 + y^2 z^2 + z^2 x^2 + 4xyz(x + y + z) \end{align} $$
Since, $x + y + z = 0 $
Therefore, we have :
$$\cfrac{x^4 + y^4 + z^4 }{2} = x^2 y^2 + y^2 z^2 + z^2 x^2 $$
EDIT :
Now, if we consider two equations : $$\begin{align} & x^2 + y^2 + z^2 = - 2(xy + yz + zx) \\ & x^3 + y^3 + z^3 = 3xyz \end{align} $$
Multiplying these two equations we get : (LHS) $$x^5 + y^5 + z^5 + x^2 y^3 + x^2 z^3 + y^2 x^3 + y^2 z^3 + z^2 x^3 + z^2 y^3 $$
And RHS as :
$$ \begin{align} & -6xyz(xy + yz + zx) \ ; \text{simplifying this further : } \\ & -6x^2 y^2 z - 6y^2 z^2 x - 6z^2 x^2 y \end{align} $$
So, this becomes :
$$x^5 + y^5 + z^5 + x^2 y^3 + x^2 z^3 + y^2 x^3 + y^2 z^3 + z^2 x^3 + z^2 y^3 = -6x^2 y^2 z - 6y^2 z^2 x - 6z^2 x^2 y $$
Therefore, we get : $$x^5 + y^5 + z^5 = -6x^2 y^2 z - 6y^2 z^2 x - 6z^2 x^2 y - x^2 y^3 - x^2 z^3 - y^2 x^3 - y^2 z^3 - z^2 x^3 - z^2 y^3 $$
EDIT -2 :
Since you have : $$ x^5 + y^5 + z^5 = -5xyz (xy + yz + zx) $$
Dividing both sides by 5 and squaring both sides : $$ \left( \cfrac{x^5 + y^5 + z^5}{5} \right)^2 = \left(-xyz (xy + yz + zx) \right) ^2 $$
$$\left( \cfrac{x^5 + y^5 + z^5 }{5} \right)^2 = x^2y^2 z^2 \left(xy + yz + zx\right)^2 $$
We also calculated $$ \cfrac{x^4 + y^4 + z^4}{2} = x^2 y^2 + y^2 z^2 + z^2 x^2 $$
Therefore, we will just put these two equations in the RHS of Required to Prove condition. $$ \begin{align} & \left( \cfrac{x^5 + y^5 + z^5 }{5} \right) \times \cfrac{x^4 +y^4 +z^4}{2} = \left(-xyz (xy + yz + zx) \right) ^2 \times x^2 y^2 + y^2 z^2 + z^2 x^2 \\ & = x^2 y^2 z^2 (x^2 y^2 + y^2 z^2 + z^2 x^2 + 2xyz(x + y + z) ) \times (x^2y^2 + y^2 z^2 + z^2 x^2) \\ & = x^2 y^2 z^2 (x^2 y^2 + y^2 z^2 + z^2 x^2 )^2 \end{align} $$
Thus, we get : $$ \left( \cfrac{x^5 + y^5 + z^5 }{5} \right) \times \cfrac{x^4 +y^4 +z^4}{2} = x^2 y^2 z^2 (x^2 y^2 + y^2 z^2 + z^2 x^2 )^2 \tag{1} $$
Now, let us try to simplify for $x^7 + y^7 + z^7 $
Consider these two equations : $$\begin{align} & x^2 + y^2 + z^2 = -2(xy + yz + zx) \\ & x^5 + y^5 + z^5 = -5xyz(xy + yz + zx) \end{align} $$
Now, multiply these 2 equations :
$$\begin{align} & \color{blue}{x^7 + y^7 + z^7} + y^2 x^5 + z^2 x^5 + x^2 z^5 + y^2 z^5 + x^2 y^5 + z^2 y^5 = 10 (x^2 y^2 + z^2 x^2 + y^2 z^2 + 2xyz(x + y + z) ) \\ & \color{blue}{x^7 + y^7 + z^7} + y^2 x^5 + z^2 x^5 + x^2 z^5 + y^2 z^5 + x^2 y^5 + z^2 y^5 = 10 (x^2 y^2 + z^2 x^2 + y^2 z^2) \\ & \color{blue}{x^7 + y^7+ z^7} + 3xyz (xy + yz + zx)^2 = 10(x^2 y^2 + z^2 x^2 + y^2 z^2) \end{align} $$
This actually comes from this :
(Dr. AKA's effort :) $$ \sum{x^2y^5+y^2x^5}=\sum{x^2y^2(x^3+y^3)}=\sum{x^2y^2(x+y)(x^2-xy+y^2)}=\sum{x^2y^2(-z)((-z)^2-3xy))}=\sum{-(xyz)^2.z+3xyz(x^2y^2)} $$ Also, $ x + y + z = 0 $ and we have : $$\begin{align} & \sum{-(xyz)^2.z+3xyz(x^2y^2)} = 3xyz\sum{x^2y^2} \\ & = 3xyz\sum{((xy+yz+zx)^2-2xyz(x+y+z))}=3xyz\sum{(xy+yz+zx)^2} \end{align} $$
So, you get : $$x^7 + y^7 + z^7 = 7xyz (x^2 y^2 + y^2 z^2 + z^2 x^2) $$
Dividing by 7 both sides and squaring : $$ \left( \cfrac{x^7 + y^7 + z^7}{7} \right)^2 = \left(xyz(x^2 y^2 + y^2 z^2 + z^2 x^2 )\right)^2 $$
Thus, we get this : $$ \left( \cfrac{x^7 + y^7 + z^7}{7} \right)^2 = x^2 y^2 z^2 (x^2 y^2 + y^2 z^2 + z^2 x^2 )^2 \tag{2} $$
Which actually concludes this :
$$\color{blue}{\left( \cfrac{x^7 + y^7 + z^7}{7} \right)^2 = \left( \cfrac{x^5 + y^5 + z^5}{5} \right)^2 \times \left( \cfrac{x^4 + y^4 + z^4}{2} \right) }$$