Roots in Progressions

Being told the roots of a cubic $x^3+px^2+qx+r=0$ are in a specific progression hands over enough extra structure to solve it, by combining the assumed form with Vieta's relations (§3.3.2.2: $\sum\alpha=-p$, $\sum\alpha\beta=q$, $\alpha\beta\gamma=-r$).

Arithmetic Progression (AP). Assume the roots are $\alpha-d,\ \alpha,\ \alpha+d$. Their sum is $3\alpha=-p$, so $\alpha=-p/3$ immediately, with no need to know $d$ yet. Since $\alpha$ itself must satisfy the cubic, substituting $x=-p/3$ into $x^3+px^2+qx+r=0$ gives a condition purely on the coefficients — for the general cubic this works out to $9pq=27r+2p^3$ (Example 3.19 pattern). Once this condition confirms an AP, find $d$ from the product relation $\alpha(\alpha-d)(\alpha+d)=\alpha(\alpha^2-d^2)=-r$.

Geometric Progression (GP). Assume the roots are $\alpha/\lambda,\ \alpha,\ \alpha\lambda$. The product relation collapses beautifully: $(\alpha/\lambda)(\alpha)(\alpha\lambda)=\alpha^3=-r$, giving $\alpha$ directly, with no need for $\lambda$ at all. Substitute $x=\alpha$ into the cubic to confirm the GP condition, then use the sum relation to solve for $\lambda$.

Harmonic Progression (HP). Roots in HP means their reciprocals are in AP. So set $y=1/x$, rewrite the equation in $y$, solve that (now an AP-roots problem, as above), then invert each $y$-root back to get $x=1/y$. …