Number of roots

Let, $y=x-a$ . Then, $y^n=y^{n-r}$ $\Rightarrow y^{n-r+r} - y^{n-r} =0$ $\Rightarrow y^{n-r} \cdot y^r-y^{n-r} =0$ $\Rightarrow y^{n-r} \cdot (y^r-1)=0$ $\Rightarrow \underbrace {y^{n-r} =0}_{1^{st}\,part}\;\,\:\text{or,}\,\;\: \underbrace {y^r=1}_{2^{nd}\,part}$

From $1^{st}$ part we get, $y^{n-r}=0\Rightarrow y=0$ $\Rightarrow x-a=0$ $\Rightarrow \underbrace {x=a}_{1 \text{ root}}$

As, $r^{th}$ root of unity will have $r$ distinct roots. From $2^{nd}$ part we get, $y^r=1$ $\Rightarrow y=x-a=\omega_1,\omega_2,\omega_3,\dots \omega_r\, \color{teal} {\text {[Here, } \omega \text{ represents the roots of unity.]}}$ $\Rightarrow x= \underbrace {(a+\omega_1) ,(a+\omega_2) ,(a+\omega_3) ,\dots, (a+\omega_r)}_{r\, \text{roots}}$

Thus, total number of distinct roots will be $\color{#0C6AC7} {\boxed {1+r}}$ .