Question

Solve ${(k-1)}^n=k^n$, where $n$ is a positive integer.

Answer 1

$(k-1{)}^n=k^n$

By expanding $(k-1{)}^n$,

$(k-1{)}^n=n_{c_0}{k}^n-n_{c_1}{k}^{n-1}+n_{c_2}{k}^{n-2}+.....+(-1{)}^n{n}_{c_n}$

$\Rightarrow (k-1{)}^n=k^n$

$\Rightarrow n_{c_0}{k}^n-n_{c_1}{k}^{n-1}+n_{c_2}{k}^{n-2}+.....+(-1{)}^n{n}_{c_n}=k^n$

$\Rightarrow k^n-n_{c_1}{k}^{n-1}+n_{c_2}{k}^{n-2}+.....+(-1{)}^n{n}_{c_n}-k^n=0$

$\Rightarrow -n_{c_1}{k}^{n-1}+n_{c_2}{k}^{n-2}+.....+(-1{)}^n{n}_{c_n}=0$

Hence, equation will be a polynomial of degree $n-1$

So there will be ${}^{\prime }n-1^{\prime }$ roots to above equation

Lets take, $(k-1{)}^n=k^n$

$\Rightarrow$ Divide both sides by $k^n$

$\Rightarrow (\frac{k-1}{x}{)}^n=1$

$\Rightarrow \frac{k-1}{k}=1^{\frac{1}{n}}$--(1)

Since, $n$-th root of unity $=e^i\frac{2p\pi }{n}$, where $p=0,1,2,....(n-1)$

$\Rightarrow \frac{k-1}{k}=e^i\frac{2p\pi }{n}$ [By using $n^{th}$ roots of unity ]

$\frac{k-1}{k}=1,\alpha ,{\alpha }^2,{\alpha }^3....{\alpha }^{n-1}$

where $\alpha =e^i\frac{2\pi }{n}$

$\Rightarrow 1-\frac{1}{k}=1,\alpha ,{\alpha }^2,...{\alpha }^{n-1}$

$\Rightarrow \frac{1}{k}=0,1-\alpha ,1-{\alpha }^2,....,1-{\alpha }^{n-1}$

Ignore $\frac{1}{k}=0$ $[\because$ we have only $n-1$ roots]

$\Rightarrow k=\frac{1}{1-\alpha },\frac{1}{1-{\alpha }^2},....,\frac{1}{1-{\alpha }^{n-1}}$

where $\alpha =e^i\frac{2\pi }{n}$