Exercise 2.2.11

(a)

$f=y^3+2y^2-3y+5=(y-\alpha )(y-\beta )(y-\gamma )=y^3-{\sigma }_1{y}^2+{\sigma }_{2}y-{\sigma }_3,$

so ${\sigma }_1=-2,{\sigma }_2=-3,{\sigma }_3=-5.$

$$\begin{array}{llll}\hfill g& =(y-\mathit{\alpha \beta })(y-\mathit{\alpha \gamma })(y-\mathit{\beta \gamma })& \hfill & \\ \hfill & =y^3-(\mathit{\alpha \beta }+\mathit{\alpha \gamma }+\mathit{\beta \gamma })y^2+({\alpha }^2\mathit{\beta \gamma }+\alpha {\beta }^2\gamma +\mathit{\alpha \beta }{\gamma }^2)y+{\alpha }^2{\beta }^2{\gamma }^2& \hfill & \\ \hfill & =y^3-{\sigma }_2{y}^2+{\sigma }_3{\sigma }_{1}y+{\sigma }_3^2& \hfill & \\ \hfill & =y^3+3y^2+10y+25.& \hfill & \end{array}$$

$y^3+3y^2+10y+25$ is the polynomial whose roots are $\mathit{\alpha \beta },\mathit{\alpha \gamma },\mathit{\beta \gamma }$.

(b)

$$\begin{array}{llll}\hfill g& =(y-\alpha -1)(y-\beta -1)(y-\gamma -1)& \hfill & \\ \hfill & =f(y-1)& \hfill & \\ \hfill & ={(y-1)}^3+2{(y-1)}^2-3(y-1)+5& \hfill & \\ \hfill & =y^3-3y^2+3y-1+2y^2-4y+2-3y+3+5& \hfill & \\ \hfill & =y^3-y^2-4y+9.& \hfill & \end{array}$$

(c)

Let $h(y)=(y-{\alpha }^2)(y-{\beta }^2)(y-{\gamma }^2)$. Then $$\begin{array}{llll}\hfill h(y^2)& =(y^2-{\alpha }^2)(y^2-{\beta }^2)(y^2-{\gamma }^2)& \hfill & \\ \hfill & =(y-\alpha )(y-\beta )(y-\gamma )(y+\alpha )(y+\beta )(y+\gamma )& \hfill & \\ \hfill & =(y^3+2y^2-3y+5)(y^3-2y^2-3y-5)& \hfill & \\ \hfill & ={(y^3-3y)}^2-{(2y^2+5)}^2& \hfill & \\ \hfill & =y^6-6y^4+9y^2-4y^4-20y^2-25& \hfill & \\ \hfill & =y^6-10y^4-11y^2-25.& \hfill & \end{array}$$

Thus

$$h(y)=(y-{\alpha }^2)(y-{\beta }^2)(y-{\gamma }^2)=y^3-10y^2-11y-25.$$

(In particular, ${\sigma }_2({\alpha }^2,{\beta }^2,{\gamma }^2)=-11$, which we can verify directly :

${\alpha }^2{\beta }^2+{\alpha }^2{\gamma }^2+{\beta }^2{\gamma }^2={\sigma }_2^2-2{\sigma }_1{\sigma }_3=9-20=-11$.)