Part 2/5:
The cubic equation in question can be structured into two components: a linear term ( py ) and a constant ( Q ). The challenge lies in resolving the ( y^3 ) term alongside the other components. To facilitate this, we can utilize V's substitution effectively. By substituting ( y ) with ( z + \frac{K}{z} ), we introduce a new variable that alters the form of the equation.
Applying the Substitution
Let’s define ( y = z + \frac{K}{z} ). Substituting this into the cubic equation results in a complex expression that needs to be expanded. Applying the binomial theorem, particularly using Pascal's triangle, allows us to expand ( (z + \frac{K}{z})^3 ) efficiently. The expansion yields terms that include ( z^3 ), ( z^2 ), and constants involving ( K ).