Find all the squareroots of x3 6x2 15x 18 0 First make i

Find all the squareroots of x^3 - 6x^2 + 15x - 18 = 0. First make it into a depressed cubic, then use Cardano\'s formula to get one squareroot. (You might need to use your calculator to get the squareroot into a recognizable form). Finally, do long division to reduce the expression to a quadratic so you can use the quadratic formula to get the last 2 squareroots.

Solution

we first eliminate the x^2 term with the substitution x = t - (1/3)(-6) = t + 2:

(t + 2)^3 - 6(t + 2)^2 + 15(t + 2) - 18 = 0
==> (t^3 + 6t^2 + 12t + 8) - (6t^2 + 24t + 24) + (15t + 30) - 18 = 0
==> t^3 + 3t - 4 = 0.

Let t = u + v. So, the cubic becomes
[u^3 + v^3 + 3uv (u + v)] + 3(u + v) - 4 = 0
==> u^3 + v^3 + (3uv + 3)(u + v) - 4 = 0.

*Now, we insist that 3uv + 3 = 0 <==> v = -1/u. This yields
u^3 + (-1/u)^3 + 0(u + v) - 4 = 0
==> u^6 - 4u^3 - 1 = 0.

We have a quadratic in u^3. Solving for u^3 yields
u^3 = 2 ± 5.
One root of this cubic is u = (2 + 5)^(1/3)

Therefore, t = u + v = u - 1/u = (2 + 5)^(1/3) - (2 + 5)^(-1/3)
==> x = t + 2 = (2 + 5)^(1/3) - (2 + 5)^(-1/3) + 2.

We can actually say more!
It can be checked that [(1 + 5)/2]^3 = 2 + 5.

Thus,
x = (2 + 5)^(1/3) - (2 + 5)^(-1/3) + 2
= (1 + 5)/2 - 2/(1 + 5) + 2
= (1 + 5)/2 - 2(1 - 5)/(1 - 5) + 2
= (1 + 5)/2 + (1 - 5)/2 + 2
= 3.