Question

If $x\ne -1$, then find the equation of $\frac{x^5+x^4+x^3+x^2}{x^3+x^2+x+1}=1$.

Answer 1

Given equation $\frac{x^5+x^4+x^3+x^2}{x^3+x^2+x+1}=1$

$\Rightarrow \frac{x^2(x^3+x^2+x+1)}{x^3+x^2+x+1}=1$

$\Rightarrow x^2=1$

$\Rightarrow x=\pm 1$

If $x=-1$, then the denominator of L.H.S will be $0,$ so, something divided by $0,$ is not defined.

$\therefore x=1$