Part 3/5:
To effectively cancel out all appearances of (y^2), we identify the contributions each expanded term provides. Upon careful arrangement and addition of the coefficients of (y^2), we ultimately derive the value of (k):
[
k = \frac{b}{3a}
]
This manipulation is crucial, as it allows all (y^2) terms to sum to zero in our transformed equation.
Moving Forward with Expanded Functions
After substituting (k) back into our cubic function, our equations begin to take on a new shape. It becomes paramount to expand these functions systematically, employing tools like Pascal's Triangle to aid in accurately managing the coefficients during binomial expansions.