Find y in terms of x of the equation: y^3 - xy = -6?

Answer 1

This can be solved using Cardano's formula for the cubic equation.

It depends on the sign of 9-x^3/27

If 9-x^3/27 >0 there is one real solution and two complex conjugate solutions as follows:

Let u =(-3 + √(9-x^3/27))^(1/3) and v =(-3 - √(9-x^3/27))^(1/3)
Then y1 = u+v
y2,3 = -(u+v)/2 ± (u-v)i√3/2


if 9-x^3/27 <0 there are three distinct real solutions. (this is known as casus irreducibilis)
if 9-x^3/27 = 0 there are three real solutions, two of which coincide.

Let r = √(x^3/27)
and cosφ = -3/√(x^3/27)

y1 = 2r^(1/3)cos(φ/3)
y2 = 2r^(1/3)cos(φ/3 + 120°)
y3 = 2r^(1/3)cos(φ/3 + 240°)

Answer 2

y^3 - xy = -6
y^3 = -6 + xy
y^2 = (-6 / y) + x
y^2 + (6 / y) = x ~~~ This is really (y^3 + 6) / y = x ~~~ But it can be expounded and the same answer will be found:
((y^2) / 6) + (1/y) = (x/6)
(y^3) / 6y) + (6 / 6y) = x/6
(y^3 + 6) / 6y = x/6
(y^3 + 6) / y = x

OR

u can...
y(y^2-x) = -6
y^2 - x = -6/y
y^2 + 6/y = x
...and go from there...

Answer 3

y^3 +6= xy
(y^3+6)/y=x
x=y^2 +6y^-1

Answer 4

y(y^2-x) = -6
y(y-x^1/2)(y+x^1/2) = -6

as far as i can go, but if you plug in numbers y = -2 and x = 1

sorry been a long while since i've messed around with trig

Answer 5

I can do x in terms of y:
y(y² - x) = -6
y² - x = -6/y
x = y² + 6/y

But you want y in terms of x...