Solutions

a) Suppose that $f(x)>0$ for $x=x_1<a$, as well as $x=x_2>a$. In b), we use the same variables $x_1$ and $x_2$. We should be able to say that $f(x)>0$ for all $x\ne a$. Indeed, if this were not so, then $f(x)\le0$ for some $x\ne a$. Clearly, $f(x)=0$ for $x\ne a$ contradicts the premise that $f(x)=0$ only for $x=a$. In addition, if $f(y)<0$ for some $y>a$, then there exists $z\ne a$ between $x_2$ and $y$ such that $f(z)=0$. The same contradiction arises if $f(y)<0$ for some $y<a$.

b) We can say that $f(x)<0$ for all $x<a$ and $f(x)>0$ for all $x>a$. For contradiction, suppose that $f(x)\le0$ for some $x>a$. Trivially, $f(x)\ne0$ for some $x>a$. Moreover, if $f(x)<0$ for some $x>a$, then there exists $z\ne a$ between $x$ and $x_2$ such that $f(z)=0$. We omit the similar proof by contradiction for the other claim that $f(x)<0$ for all $x<a$.

c) Notice that we can factor the formula:

$$\begin{align*} x^3+x^2y+xy^2+y^3 &= x^2(x+y)+y^2(x+y)\\ &=(x+y)(x^2+y^2). \end{align*}$$

If both $x$ and $y$ have the same sign, then the formula's sign is their sign. Otherwise, the sign of the variable whose absolute value is greatest becomes the sign of the formula.