If one root of the equation `ax^2 + bx + c = 0` is equal to the `n^(th)` power of the other, then `(ac^n)^(1/(n+1)) + (a^nc)^(1/(n+1)) + b` is equal to

Explanation

Let `alpha` be one root of the equation `ax^(2)+bx+c=0`
then other root be `alpha^(n)`
`:.alpha +alpha^(n)=-b/a` ……..i
and `alpha. alpha^(n)=c/a`
`impliesalpha^(n+1)=c/a`
`impliesalpha=(c/a)^(1/(n+1))`
`:.` From Eq. (i) we get
`(c/a)^(1/(n+1))+(c/a)^(n/(n+1))=-b/a`
`implies(c)^(1/(n+1)).a^(-1/(n+1)+1)+(c^(n))^(1/(n+1)).a^(-n/(n+1)+1)+b=0`
`implies1/c^(1/(n+1)).a^(n/(n+1))+(c^(n))^(1/(n+1)).a^(1/(n+1))+b=0`
`=(a^(n)c)^(1/(n+1))+(c^(n)a)^(1/(n+1)+b=0`