When is one of the roots of a cubic equal to the product of the other two roots?

This is not easy at all. The condition (one of several that will do) is:

a = (1/2d)(-c² + 2bd + 2cd - d² ± √ (-4 b²d² + (c² - 2bd - 2cd + d²)² ) )

I'll wait and see if anyone can improve on this.

Edit: A much more concise expression can be had if the cubic is first reduced to the form:

x³ + ax² + bx + c = 0

Then the necessary and sufficient condition is just

a = 1 + i (b/√c - √c) or 1 - i (b/√c - √c)

Yes, that i has to be there. This solution has a similar form to that of gôhpihán's.

One of +/- sqrt(-d/a) is a root. Additionally, if d=0, 0 is (at least) a double root.

Say the roots are u, v, and w. Of course, -d/a = uvw.

Necessary:

Suppose uv = w. Then -d/a = w^2, so w = +/- sqrt(-d/a) is a root. If d=0, then w = 0 = uv, so u=0 or v=0, so 0 is (at least) a double root.

Sufficient:

Suppose w = +/- sqrt(-d/a) is a root. Then w^2 = -d/a = uvw. If w != 0, then w = uv, as needed. If w=0, then by assumption 0 is a double root, so we may say v=0, giving w = uv (0 = 0).

I hope I'm not missing some complexity to this question; it is a bit late....

Edit: I did miss something: this is basically Suleiman's method. I can console myself with thinking my writeup is a little neater, I suppose.

Dividing both sides of the equation by a will preserve its roots,

so assume WLOG that a = 1.

Set f(x) = x³ + bx² + cx + d.

Case 1: d = 0.

Claim: one root is a product of the others iff c = 0.

Proof.

If we have f(x) = x³ + bx = x²(x + b) then the roots are 0 with multiplicity 2, and -b,

and of course 0 = 0(-b).

On the other hand, d = 0 implies that x = 0 is a root, so clearly one root is the product of the others iff x = 0 has multiplicity 2. So for some p,

f(x) = x²(x - p) = x³ - px².

Thus c = 0.

Case 2: d ≠ 0.

Claim: one root is the product of the others iff a square root of -d is a root.

Proof.

Suppose f(√(-d)) = 0 (the case f(-√(-d)) = 0 is the same). Then for some p and q

f(x) = (x - p)(x - q)(x - √(-d))

= x³ - (p + q + √(-d))x² + (pq + p√(-d) + q√(-d))x - pq√(-d).

But then d = -pq√(-d) ⇒ √(-d) = pq.

So √(-d) is the product of the other two (unknown) roots.

Conversely, suppose one root of f is the product of the other two, i.e. for some α,β

f(x) = (x - α)(x - β)(x - αβ)

= x³ - (α + β + αβ)x² + (α²β + αβ² + αβ)x - α²β².

Then we have that √(-d) = √(α²β²) = αβ which is a root.

Conclusion:

For one root of f to be the product of the other two, it is necessary and sufficient for one of c, f(√(-d)), and f(-√(-d)) to be zero.

ax^3 + bx^2 + cx + d = 0

Let e, f, g be the roots of the cubic equation

Then by Vieta's Formula

e + f + g = -b/a ............. (1)

ef + eg + fg = c/a ............ (2)

efg = -d/a .................. (3)

Given

e = fg

(1):

fg + f + g = -b/a

fg + f + g + 1 = -b/a + 1

(2):

ef + eg + e = c/a

e(f + g + 1) = c/a

e(-b/a + 1 - fg) = c/a

-be/a + e - efg = c/a

-be/a + e + d/a = c/a

-be + e + d = c

e(-b + 1) = c - d

e = (d - c)/(b - 1)

e² = ( (d - c)/(b - 1) )²

(3):

e² = -d/a

Thus the a necessary condition is that

( (d - c)/(b - 1) )² = -d/a

OR

d(b - 1)² + a(d - c)² = 0

I think I'm missing something here again....

x^3 - 3x^2 + 3x - 1 <=== probably the most simple ...

(x - 1)^3 ... all three roots = 1 ... 1 = 1 * 1

********

x^3 - 11x^2 + 36x - 36

(x - 6) * (x - 2) * (x - 3)

roots of 2, 3, and 6 ... 2 * 3 = 6