Please forgive me for any mistake in the proposal of this problem
Problem
The first step of Cardano's Method states:-Remove the term containing $x^2$ by taking $y=x-\dfrac{b}{3a}$. However if we are not able to remove the $x^2$ in a equation then how will we solve it.
Any help is welcome. Thanks in advance
Your question doesn't make sense. If, in $ax^3+bx^2+cx+d$ (with $a\neq0$), $x$ gets replaced by $y-\frac b{3a}$ and then you expand it, you always get a cubic without a monomial with degree $2$, namely$$ay^3+\left(c-\frac{b^2}{3a}\right)y+\frac{2 b^3}{27 a^2}-\frac{c b}{3 a}+d.$$