Prove that for a cubic of the form $\displaystyle a(x-b)^3+d$ then there exists at least one set of real solutions $\displaystyle \{a,b,d\}$ such that:

$\displaystyle y_1=a(x_1-b)^3+d$

$\displaystyle y_2=a(x_2-b)^3+d$

$\displaystyle y'_1 = 3a(x_1-b)^2$