x^n + y^n is divisible by | vedclass

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(C) The expression $x^n + y^n$ is divisible by $(x + y)$ if and only if $n$ is an odd positive integer.For $n = 1$, $x^1 + y^1 = x + y$, which is divi

Vedclass · Accepted Answer

$x + y$ for all $n = 2m - 1, m \in N$

Vedclass · Answer

(C) The expression $x^n + y^n$ is divisible by $(x + y)$ if and only if $n$ is an odd positive integer.For $n = 1$, $x^1 + y^1 = x + y$, which is divisible by $(x + y)$.For $n = 2$, $x^2 + y^2$ is not divisible by $(x + y)$.For $n = 3$, $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$, which is divisible by $(x + y)$.In general, for any odd integer $n = 2m - 1$ where $m \in N$, $x^n + y^n$ is divisible by $(x + y)$.

$x^n + y^n$ is divisible by

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