Solving general general cubic and quartic polynomials by radicals is a problem with an interesting history.

Anyways, if you want to solve your numerically, you can use Newton's method to get a real root:

x ≈ 2.72741

Factoring x - 2.72741... out of your cubic x^{3 }+ x^{2 }- x - 25 = 0 gives a quadratic, which when set to 0 and solved by quadratic formula gives the other two roots, which are complex:

x ≈ -1.86371 ± 2.38596 i

If yo want to solve your cubic by radicals, you can do it as follows, since it fails the rational root test.

Step 1:

In the cubic ax^{3}+bx^{2}+cx+d=0, if you substitute x = y - b/(3a), expand and simplify by collecting terms with equal powers of y, you get a cubic polynomial in y in which there is no y^{2} term. We also need to divide both sides by a and put the linear and constant term on the right hand side. The resulting cubic is called a **depressed** cubic (general form is y^{3} = py + q).

With your cubic, x^{3} + x^{2} - x - 25 = 0, we let x = y - 1/(3*1) = y - 1/3 and substitute.

(y - 1/3)^{3} + (y - 1/3)^{2} - (y - 1/3) - 25 = 0

y^{3} - y^{2} + (1/3)y - 1/27 + y^{2} - (2/3)y + 1/9 - y + 1/3 - 25 = 0

y^{3} - (4/3)y - 664/27 = 0

y^{3} = (4/3)y + 664/27 (This is our depressed cubic equation).

Step 2: Solving the depressed cubic.

Observe the identity (s+t)^{3} = 3st(s+t) + s^{3}+t^{3}. I will leave it to you as an exercise to verify this identity.

Notice that it is of the form y^{3} = py + q where p = 3st, q = s^{3} + t^{3}, and y =s+t.

Thus if we find two numbers p = 3st, q = s^{3} + t^{3}, we will find that y = s + t is a root.

If we factor out y-(s+t) and solve the resulting quadratic by the quadratic formula, it turns out that the other two roots are

y = -(s+t)/2 ± (√3)(s-t)i/2.

With your cubic, for example, we derived the depressed cubic and saw that p = 4/3 and q = 664/27 so we seek numbers s and t satisfying 3st = 4/3 and s^{3} + t^{3} = 664/27.

In the first equation, dividing by 3 and cubing gives s^{3}t^{3} = 64/729. Thus s^{3} and t^{3} are roots of a quadratic

u^{2} - (664/27)u +64/729 = 0

which when solved using the quadratic formula gives u = (332 ± 36√85)/27.

Taking cube roots give the expressions for s and t so we can back substitute in the formulas for y and then back substitute the y values to get x. Here is the final result:

x = (-1 + ^{3}√(332 + 36√85) + ^{3}√(332 - 36√85) )/3

x = -1/3 - (^{3}√(332 + 36√85) + ^{3}√(332 - 36√85))/6 + i(√3)(^{3}√(332 + 36√85) - ^{3}√(332 - 36√85))/6

x = -1/3 - (^{3}√(332 + 36√85) + ^{3}√(332 - 36√85))/6 - i(√3)(^{3}√(332 + 36√85) - ^{3}√(332 - 36√85))/6

Land L.

Thank you for introducing the "depressed cubic" method. It really helps me a lot in solving cubic equation.

However, in your method there can only be a solution in the step of taking p = 3st, q = s

^{3}+ t^{3}, if there is really 3 real roots, can this method also solve for all of them? If yes, how should it be done? If not, what will be the root solved out? The largest or the smallest?I would be thankful if you can answer my problem. Sorry for my bad grammar.

01/13/13