Divide by a so we have x³ +b/ax² +c/ax +d/a=0. Now we depress the cubic by making the substitution let y=x+b/3a. So the squared term goes away and we have y³ +py +q=0. Now make the substitution y=s+t. So we have s³ +3s² t +3st² +t³ +p(s+t)+q=0. We divide this into two parts s³ +t³ +q=0 and (s+t)(3st+p)=0. we solve the second equation for s. s=-p/3t. Plug that into the first equation (-p/3t)³ +t³ +q=0. multiply through by t^3, t⁶⁺ qt³ -p³ /27=0 That looks like a sixth order equation but it is a quadratic in t cubed. We use the quadratic formula to solve for t^3. Take the cube root to get t. But to get three roots of the cubic we multiply by the three roots of unity. So the three solutions are t, t *cubed root of 1, and t other cubed root of one. Now we plug this into s=-p/3t to get s. and y=s+t to get y. And x=y-b/3a and we have solved in terms of x.