Hey David, here are some hints for problems 26 - 29:
26: Use substitution. Note the left side of the equation is $x + xy + y = (x+y) + xy$ and the right side of the equation is $x^2 + xy +y^2 = (x+y)^2 -xy$.
27: Part (a) says we can think of $x^3 + y^3 + z^3 -3xyz$ as a polynomial in $z$, that is, think of $x$ and $y$ as constants. So, for part (b) you want to plug in $z = -(x+y)$ and verify that the expression you get is equal to $0$. Then, in (c), since $z = -(x+y)$ is a root of the polynomial, it must be true that $z - (-(x+y)) = x + y + z$ is a factor of $x^3 + y^3 + z^3 -3xyz$, so you can use long division to find the other factor.
28: Try using the general version of Vieta's Theorem. For example, you can start by noting that $t = -(a+b)(b+c)(c+a)$ since $a+b$, $b+c$, and $c+a$ are the roots of the second polynomial.
29: Try rewriting the expression using $u = x+y$ and $v = xy$.
Vieta's formulas are true regardless if the roots are real or imaginary. You do not want to find the roots, you just want to find the value of $t$, so there's no need to worry if the roots are real or imaginary.