Expressions Relating to Roots
Once you can build α + β and αβ, almost any symmetric combination of roots — sums of cubes, sums of reciprocals, products of pairs — falls out with the right identity.
Building more complex symmetric expressions
Any expression that is symmetric in the roots (unchanged when the roots are permuted among themselves) can always be rewritten in terms of the elementary symmetric sums given by Vieta's formulae. The standard toolkit of identities includes:
$$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)$$
$$\frac1\alpha+\frac1\beta+\frac1\gamma=\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}$$
$$\frac1\alpha+\frac1\beta+\frac1\gamma+\frac1\delta$$
$$=\frac{\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta}{\alpha\beta\gamma\delta}$$
$$\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma=(\alpha+\beta+\gamma)\big(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha\big)$$
The reciprocal identity extends the same way to a quartic: the numerator is always the sum of products of the roots taken three at a time (i.e. "all roots except one, summed over every choice of the missing root"), and the denominator is the product of all the roots.
$\alpha+\beta+\gamma=4$, $\alpha\beta+\beta\gamma+\gamma\alpha=5$, $\alpha\beta\gamma=2$.
$\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma=\dfrac{5}{2}$.
$\alpha+\beta+\gamma=-1$, $\alpha\beta+\beta\gamma+\gamma\alpha=-4$, $\alpha\beta\gamma=4$.
$\alpha^2+\beta^2+\gamma^2=(-1)^2-2(-4)=1+8=9$.
$\alpha^3+\beta^3+\gamma^3-3(4)=(-1)\big(9-(-4)\big)=(-1)(13)=-13$.
$\alpha^3+\beta^3+\gamma^3=-13+12=-1$.
Products and sums of pairs
$\alpha+\beta+\gamma=\dfrac32$, $\alpha\beta+\beta\gamma+\gamma\alpha=-\dfrac{11}{2}$, $\alpha\beta\gamma=-3$.
Key trick: $\alpha+\beta=(\alpha+\beta+\gamma)-\gamma$, so $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$ can be rewritten using $s=\alpha+\beta+\gamma$ as $(s-\gamma)(s-\alpha)(s-\beta)$.
This equals $s^3-s^2(\alpha+\beta+\gamma)+s(\alpha\beta+\beta\gamma+\gamma\alpha)-\alpha\beta\gamma=s^3-s^3+s(\alpha\beta+\beta\gamma+\gamma\alpha)-\alpha\beta\gamma$ $=s(\alpha\beta+\beta\gamma+\gamma\alpha)-\alpha\beta\gamma$.
$=\dfrac32\left(-\dfrac{11}{2}\right)-(-3)=-\dfrac{33}{4}+3=-\dfrac{33}{4}+\dfrac{12}{4}=-\dfrac{21}{4}$.
Whenever an expression involves $(\alpha+\beta)$, $(\beta+\gamma)$, $(\gamma+\alpha)$, it's almost always fastest to substitute $\alpha+\beta=s-\gamma$ (where $s=\alpha+\beta+\gamma$) and similarly for the others, turning the product into a symmetric function of $\alpha,\beta,\gamma$ directly, which then telescopes using the general identity $(s-\alpha)(s-\beta)(s-\gamma)=s^3-s^2\Sigma_1+s\Sigma_2-\Sigma_3$ evaluated at $s=\Sigma_1$.
Note there is no $x^3$ term, so $\alpha+\beta+\gamma+\delta=0$, and no $x$ term, so $\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=0$.
Coefficient of $x^2$ gives $\sum\alpha\beta=-5$ (all six pairwise products).
$\alpha^2+\beta^2+\gamma^2+\delta^2=(\alpha+\beta+\gamma+\delta)^2-2\sum\alpha\beta=0^2-2(-5)=10$.
When a term is "missing" from a polynomial (e.g. no $x^3$ term in a quartic), the corresponding symmetric sum is zero, not undefined or omitted from calculations — always read off every coefficient, including implicit zero ones.
Shifted products of roots
An expression like $(\alpha+k)(\beta+k)(\gamma+k)$, for some constant $k$, is also symmetric in the roots, so it too reduces to the elementary sums once expanded fully as a polynomial in $k$.
$\alpha+\beta+\gamma=-3$, $\alpha\beta+\beta\gamma+\gamma\alpha=-2$, $\alpha\beta\gamma=-8$.
Expand the product fully: $(\alpha+2)(\beta+2)(\gamma+2)=\alpha\beta\gamma+2(\alpha\beta+\beta\gamma+\gamma\alpha)+4(\alpha+\beta+\gamma)+8$.
$=-8+2(-2)+4(-3)+8=-8-4-12+8=-16$.
For $(\alpha-k)(\beta-k)(\gamma-k)$ the same expansion applies with $k$ replaced by $-k$ throughout — it's often quickest to expand $(\alpha+k)(\beta+k)(\gamma+k)$ generally once, memorise the pattern (constant term $\alpha\beta\gamma$, then coefficient $k$ times the pair-sum, then $k^2$ times the root-sum, then $k^3$), and substitute the correct sign of $k$ each time rather than re-deriving from scratch.
Exam-style questions
Question 1 5 marks
The roots of $x^3-3x^2+2x-6=0$ are $\alpha,\beta,\gamma$. Find $\dfrac{1}{\alpha\beta}+\dfrac{1}{\beta\gamma}+\dfrac{1}{\gamma\alpha}$.
$\alpha+\beta+\gamma=3$, $\alpha\beta+\beta\gamma+\gamma\alpha=2$, $\alpha\beta\gamma=6$ (B1)
$\dfrac{1}{\alpha\beta}+\dfrac{1}{\beta\gamma}+\dfrac{1}{\gamma\alpha}=\dfrac{\gamma+\alpha+\beta}{\alpha\beta\gamma}=\dfrac{3}{6}=\dfrac12$ (M1, A1, A1, A1)
Question 2 6 marks
The roots of $x^3+2x^2-3x+5=0$ are $\alpha,\beta,\gamma$. Find $\alpha^3+\beta^3+\gamma^3$.
$\alpha+\beta+\gamma=-2$, $\alpha\beta+\beta\gamma+\gamma\alpha=-3$, $\alpha\beta\gamma=-5$ (B1)
$\alpha^2+\beta^2+\gamma^2=(-2)^2-2(-3)=4+6=10$ (M1, A1)
$\alpha^3+\beta^3+\gamma^3-3(-5)=(-2)(10-(-3))=(-2)(13)=-26$ (M1, A1)
$\alpha^3+\beta^3+\gamma^3=-26-15=-41$ (A1)
Question 3 5 marks
The roots of $2x^3-x^2-7x+6=0$ are $\alpha,\beta,\gamma$. Find $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$.
$\alpha+\beta+\gamma=\dfrac12$, $\alpha\beta+\beta\gamma+\gamma\alpha=-\dfrac72$, $\alpha\beta\gamma=-3$ (B1)
$(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)=(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)-\alpha\beta\gamma$ (M1)
$=\dfrac12\left(-\dfrac72\right)-(-3)=-\dfrac74+3=\dfrac{5}{4}$ (M1, A1, A1)