[tex] \displaystyle a)~f(1)=n \cdot 1^n+1-n \cdot 1-1=n+1-n-1=0. \\ \\ b)~Daca~n~este~impar,~rezulta~(-1)^n=-1.\\ \\ f(-1)=n \cdot (-1)^n+(-1)^2-n \cdot (-1)-1=-n+1+n-1=0.\\ \\ Din~f(1)=0~si~f(-1)=0~rezulta~X-1~|~f~si~X+1~|~f,~deci \\ \\ (X-1)(X+1)~|~f~\Leftrightarrow X^2-1~|~f. [/tex]
[tex] \displaystyle c)~ Presupunem~ca~ar~exista~o~radacina~c \in \mathbb{Q}- \mathbb{Z}. \\ \\ Atunci~c= \frac{p}{q},~unde~ p,q \in \mathbb{Z},~q \neq 0~si~(p,q)=1. \\ \\ f(c)=0 \Leftrightarrow n \cdot \left( \frac{p}{q} \right)^n+ \left( \frac{p}{q} \right)^2-n \cdot \frac{p}{q}-1=0.\\ \\ Inmultim~cu~q^n \Rightarrow np^n+p^2q^{n-2}-npq^{n-1}-q^n=0 \Leftrightarrow \\ \\ \Leftrightarrow np^n=q^n+npq^{n-1}-p^2q^{n-2} \Leftrightarrow \\ \\ \Leftrightarrow np^n=q^{n-2}(q^2+npq-p^2). [/tex]
[tex] \displaystyle Acum~observam~ca~membrul~stang~este~divizibil~cu~p. \\ \\ Rezulta~ca~si~membrul~drept~este~divizibil~cu~p. \\ \\Deci~p~|~q^{n-2}(q^2+npq-p^2). \\ \\ Cum~(p,q)=1,~rezulta~p~|~q^2+npq-p^2. \\ \\ Observam~insa~ca~q^2+npq-p^2=q^2+p(nq-p). \\ \\ Cum~p~|~q^2+p(nq-p)~si~p~|~p(nq-p),~rezulta~ca~p~|~q^2.~Dar~cum \\ \\ (p,q)=1,~acest~lucru~este~posibil~doar~daca~p=q=1. [/tex]
[tex] \displaystyle Deci~c=1 \notin \mathbb{Q}- \mathbb{Z},~contradictie! [/tex]
