Prove that $f=(x+i)^{10}+(x-i)^{10}$ have all real roots We have $f=(x+i)^{10}+(x-i)^{10}$ and we need to prove that $f$ have all the roots in $\mathbb{R}$.
Here is all my steps:


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*Suppose that $z\in\mathbb{R}$ is a root of $f\Rightarrow (z+i)^{10}+(z-i)^{10}=0$ 


Therefore: $f(z)=\sum_{k=0}^{10}\left[\left(\dbinom{10}{k}\cdot z^{10-k}\cdot i^k\right)\left(1+(-1)^k\right)\right]=0$

I don't have ideea how can I prove that $f$ have all real roots

Suppose that $z=a+bi$ and we need to prove that $b=0$:
$$\Rightarrow f(z)=\sum_{k=0}^{10}\left(\dbinom{10}{k}\cdot a^{10-k}\cdot i^k\right) \left[(b+1)^k+(b-1)^k \right]=0$$
$$\Rightarrow \left[(b+1)^k+(b-1)^k \right]=0$$
$$\Rightarrow b+1=-b+1$$
$$\Rightarrow b=0$$
Therefore $f$ have all the roots in $\mathbb{R}$


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*Here is a photo with the proof of real axis, I don't know if is correct: 




$x=a+bi$ 
$$(x+i)^{10}+(x-i)^{10}=0 \Rightarrow (x+i)^{10}=-(x-i)^{10} \Rightarrow (\frac{x+i}{x-i})^{10}=-1 \\ \Rightarrow \left (\frac{a+(b+1)i}{a+(b-1)i}\right )^{10}=i^{10} \Rightarrow \frac{a+(b+1)i}{a+(b-1)i}=i \\ \Rightarrow \frac{(a+(b+1)i)(a-(b-1)i)}{(a+(b-1)i)(a-(b-1)i)}=i \Rightarrow \frac{a^2-a(b-1)i+a(b+1)i+(b^2-1)}{a^2+(b-1)^2}=i \Rightarrow \frac{(a^2+(b^2-1))+2ai}{a^2+(b-1)^2}=i$$ 
That means that $$\frac{a^2+(b^2-1)}{a^2+(b-1)^2}=0 \text{ and } \frac{2a}{a^2+(b-1)^2}=1$$ 
$$\frac{a^2+(b^2-1)}{a^2+(b-1)^2}=0 \Rightarrow a^2+b^2-1=0 \Rightarrow b^2=1-a^2$$ 
$$\frac{2a}{a^2+(b-1)^2}=1 \Rightarrow 2a=a^2+(b-1)^2 \Rightarrow 2a=a^2+b^2-2b+1 $$ 
Solving for $a$ and $b$ we get the following solutions: 
$a=0, b=1$ and $a=1, b=0$ 
We reject $(a, b)=(0, 1)$ because then the denominator would be $0$. 
So, the solution is $(a, b)=(1, 0)$. 
That means that $x=1 \in \mathbb{R}$. 
$$$$ 
Do the same for the case $\left (\frac{a+(b+1)i}{a+(b-1)i}\right )^{10}=i^{10} \Rightarrow \frac{a+(b+1)i}{a+(b-1)i}=-i$
