If $\alpha ,\beta $ are roots of the equation $2x^{2}+6x+b=0$ where $b<0$, then find least integral value of $\displaystyle \left ( \dfrac{\alpha ^{2}}{\beta }+\dfrac{\beta ^{2}}{\alpha } \right )$.

alpha + beta = -3, alpha*beta = b/2. Expression = (alpha^3 + beta^3) / (alpha*beta) = ((alpha+beta)^3 - 3*alpha*beta*(alpha+beta)) / (alpha*beta) = (-27 - 3*(b/2)*(-3)) / (b/2) = (-27 + 4.5b) / (b/2) = -54/b + 9. Since b < 0, let b = -k (k>0). Expression = 54/k + 9. To minimize this, k should be large, but b is a coefficient. With integer roots or constraints, 10 is the least value.