urgently need the solution?

Question

solve
36 {y ^ (3)} - 3 {y ^ (2)} = b ^ (2) - 8a. here b and a are constants
for e.g.36 - 3 = 81 - 48
here,y=1 when b = 9 and a = 6
this i found at random pls solve it for y in general.
the best answerer will surely get 10 points and
lot of my regards.

Answer 1

I assume you are looking for integer solutions only. Here is another one: y = 2, b = 18, a = 6.

Here's how I found it. I set y = 2 and evaluated the left side of the equation which came to 276. Then on the right side, I tried a = 1, but that didn't give a square number for b^2, so I kept incrementing a by one until I had a square number (324) for b^2.

It wasn't very hard. You could probably write a computer program to find more solutions. I'm not positive, but I think you could probably find an integer a and b for any integer y.

By the way, this looks like a Diophantine problem. You can read more about them at http://en.wikipedia.org/wiki/Diophantine_equation

P.S. If the variables don't have to be integers, you can just pick anything you want for two of them and solve for the third one.

Answer 2

this is the same as 36 {y ^ (3)} - 3 {y ^ (2)} = c where c = b^2 - 8a

or y^3 - (1/12)y^2 + d =0 where d = -c/36

put this into reduced form by y = x+1/36

x^3 - (1/432)x - (1/23328 + d) = 0

Then use Cardano's formula.

the first root of the reduced form is

x= {-q/2 + √[q^2/4 + p^3/27]}^(1/3) + {-q/2 - √[q^2/4 + p^3/27]}^(1/3)

where p = -1/432 and q = - (1/23328 + d)

then y = x+1/36.

Answer 3

The equation u have shown definitely have 3 solutions for y but as far as various values of a and b are concerned u'll have infinite solutions....and be careful with the term constant if a and b are constants then how can u randomly select it's values...This will apear vague to u till u learn how to find a solution for a polynomial of degree 3...

Answer 4

huh?