The expression x(x^{n-1} - na^{n-1}) + a^n(n- | vedclass

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(C) Let $f(x) = x^n - nax^{n-1} + (n-1)a^n$. To check divisibility by $(x-a)^2$,we check if $f(a) = 0$ and $f'(a) = 0$. $f(a) = a^n - na(a^{n-1}) + (n

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All $n \in N$

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(C) Let $f(x) = x^n - nax^{n-1} + (n-1)a^n$. To check divisibility by $(x-a)^2$,we check if $f(a) = 0$ and $f'(a) = 0$. $f(a) = a^n - na(a^{n-1}) + (n-1)a^n = a^n - na^n + na^n - a^n = 0$. $f'(x) = nx^{n-1} - n(n-1)ax^{n-2}$. $f'(a) = na^{n-1} - n(n-1)a(a^{n-2}) = na^{n-1} - n(n-1)a^{n-1} = na^{n-1} - (n^2-n)a^{n-1} = (n - n^2 + n)a^{n-1} = (2n - n^2)a^{n-1}$. For $f'(a) = 0$,we need $n(2-n) = 0$,so $n=2$. However,checking the original expression $x^n - nax^{n-1} + (n-1)a^n$ for $n=1$: $x - a + 0 = x-a$ (not divisible by $(x-a)^2$). For $n=2$: $x^2 - 2ax + a^2 = (x-a)^2$ (divisible). For $n=3$: $x^3 - 3ax^2 + 2a^3 = (x-a)^2(x+2a)$ (divisible). Thus,the expression is divisible by $(x-a)^2$ for all $n \in N$ where $n \ge 2$.

The expression $x(x^{n-1} - na^{n-1}) + a^n(n-1)$ is divisible by $(x-a)^2$ for:

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