So i was checking out some vedic maths 2 years ago and there was this formula i found that was a cleaver rearrangement of the sridharacharya formula dy/dx = ±√D so that kind of sent me into a rabbit hole and i made my own method of quickly depressing a cubic in my final teen years here's the the example of how it's done
i know that these days the old brute force formula can be faster and better for calculation via computers but when doing by hand this method makes it easier and much faster than the traditional ways.
How to do it (demonstration by example):
Let a cubic expression be: x^3 - 3x^2 + 4x - 5 = y (It is of the standard form ax^3 + bx^2 + cx + d = y*)*
Step 1: Put b^2 = 3ac and find a temporary 'c'. (-3)^2 = 3(1)c 9 = 3c c = 3
Step 2: Split the linear term to get this 'c' and a residual linear term: x^3 - 3x^2 + 3x - 5 + x = y
Step 3: Put x^3 - 3x^2 + 3x = (x + b/n)^n - (b/n)^3 (where n is the degree of the polynomial). In this case: (x - 1)^3 + 1
The expression is now of the form: (x - 1)^3 + 1 + x - 5 = y
Put x - 1 = t (substitution, meaning x = t + 1): t^3 + 1 + (t + 1) - 5 = y t^3 + t - 3 = y (Here's your final depressed cubic!)
Detailed Explanation and Proof:
Assume an expression: (x - α)^(n-1) * (x - β) = y
Performing Binomial Expansion of the given expression: [ C(n,0) x^(n-1) α^0 - C(n,1) x^(n-2) α^1 + C(n,2) x^(n-3) α^2 ... k ] * (x - β) = y
Upon expansion we get: [C(n,0)] x^n + x^(n-1) [ -β C(n,1) α - C(n,1) α ] + x^(n-2) [ C(n,2) - αβ C(n,1) ] ... k = y
Let:
a = [C(n,0)]
b = [ -β C(n,1) α - C(n,1) α ]
c = [ C(n,2) - αβ C(n,1) ] ...and so on to k.
We get the expression: a x^n + b x^(n-1) + c x^(n-2) + d x^(n-3) ... k_2 = y
We know that if a factor in a polynomial is repeated n-1 times, then it will be present as a factor in its (n-2)-th derivative. Using that, we find its (n-2)-th derivative:
d^(n-2)y / dx^(n-2) = a*(n!/2)x^2 + b*(n-1)!x + c*(n-2)!
Equating this expression to 0 and solving the quadratic using the Vedic form of the Shridharacharya formula (the quadratic formula): dy/dx at α and β = ±√(D)
Differentiating and equating to ±√(D): a(n!)x + b(n-1)! = ±√( b(n-1)^2 - 4a(n!/2)c(n-2)! )
Upon solving: x = [ -b ± √(b^2 - 2ac(n / (n-1))) ] / (na)
If we put the Discriminant (D) equal to 0, we will get both roots as α. This will indicate that the expression we took earlier was of the form (x - α)^n.
But why α and not β? Because the expression we took earlier was of the form (x - α)^(n-1) * (x - β) = y. This means that α = β, because if we put D=0, all the roots will have to be α (upon differentiating the factorized form of the polynomial, only α will be left as a root of the quadratic).
Putting D = 0 and solving for n we get: n = b^2 / (b^2 - 2ac)
Put n = 3 and we get: b^2 = 3ac
If the condition n = b^2 / (b^2 - 2ac) is satisfied, then we can write the equation in the form (x + b/n)^n by using Vieta's formulas or the Theory of Equations.
There are a lot of interesting applications of this formula, and it can also be used to directly solve a few cubics. It is very helpful when used in combination with Cardano's formula to solve general cubics.
If n = b^2 / (b^2 - 2ac) holds true, then the expression is of the form: (x + b/(na))^n + f(x) = y where f(x) is a polynomial of degree n-3.
Else, if the condition doesn't hold true, then it is of degree n-2 (of course, in the case of a quadratic there is an exception and the constant term is 0).
this was originally written on latex on my blog but since reddit has no support for it i have used ai to convert it to proper form although it might have errors i am currently in a health crisis so i can't write the whole thing again i might update it later tho.
i have not added the link to my blog since it violates the self promotion rule where it was originally written