Solve$18x^3+81x^2+121x+60=0$, one root being half the sum of the other two.

: Let the roots be $\alpha ,\beta ,\gamma$

Given that, $\alpha =\frac{1}{2}(\beta +\gamma )$

$\begin{array}{l}2\alpha =\beta +\gamma \\ \alpha +\beta +\gamma =-\frac{81}{18}\end{array}$

$\alpha +2\alpha =-\frac{81}{18}\Rightarrow 3\alpha =-\frac{81}{18}\Rightarrow \alpha =-\frac{27}{18}$

$\beta +\gamma =2(-\frac{27}{18})=-\frac{27}{9}\Rightarrow \gamma =-\frac{27}{9}-\beta$

$\alpha \beta \gamma =-\frac{60}{18}$

$-\frac{27}{18}\left(\beta \gamma \right)=-\frac{60}{18}\Rightarrow \beta \gamma =\frac{60}{27}=\frac{20}{9}$

$\beta (-\frac{27}{9}-\beta )=\frac{20}{9}$

$\beta \frac{(-27-9\beta )}{9}=\frac{20}{9}$

$-9{\beta }^2-27\beta -20=0$

$9{\beta }^2+27\beta +20=0$

$9{\beta }^2+15\beta +12\beta +20=0$

$3\beta (3\beta +5)+4(3\beta +5)=0$

$(3\beta +5)(3\beta +4)=0$

$\beta =-\frac{5}{3},\beta =-\frac{4}{3}$

$\text{Case}\left(i\right):\alpha =-\frac{27}{18},\beta =-\frac{5}{3}$

$\gamma =-\frac{27}{9}+\frac{5}{3}=\frac{-27+15}{9}=\frac{-12}{9}=-\frac{4}{3}$

$\therefore$Roots are$-\frac{27}{18},-\frac{5}{3},-\frac{4}{3}$

$\text{Case}\left(ii\right):\alpha =-\frac{27}{18},\beta =-\frac{4}{3}$

$\gamma =-\frac{27}{9}+\frac{4}{3}=\frac{-27+12}{9}=\frac{-15}{9}=-\frac{5}{3}$

$\therefore$Roots are $-\frac{27}{15},-\frac{4}{3},-\frac{5}{3}$