ax

^{2}+ bx + c = 0

which are then solved, either by factoring and solving for x, or - very often - by using the quadratic formula:

x = [-b

__+__{b

^{2 }- 4ac}

^{1/2}]/ 2a

But what about solving basic cubic equations? Here we solve equations of the form:

x

^{3}– px

^{2}+ qx – r = 0

**Example:**

x

^{3}– x

^{2}- x + 2 = 0

What I’ll show here, or rather demonstrate, is an algorithm for simple solution provided the numerical coefficient of the cubic term is 1.

We begin by writing: p = 1, q = -1, r = -2

And let:

D = [(- 4p

^{3 }r – 27r

^{2}+ 18 pqr – 4 q

^{3}+ p

^{2}q

^{2})]

^{1/2}

With:

x1 = [p

^{3}–9/2 (pq – 3r) – 3/2 x (-3)

^{1/2}x D ]

^{1/3}

x2 = (p

^{2}– 3q)/ x1

The ROOTS are then – in turn:

a1 = 1/3(p + r2 x x1 + r x x2)

WHERE:

r = - 1/2 + (-3)

^{1/2}/2 x (-3)

^{1/2}D

r2 = -1/2 – (-3)

^{1/2}/2

Last TWO roots:

a2 = 1/3( p + r2 x x1+ r x x2)

a3 = 1/3( p + r x x1 + rx r2 x x2)

For the original eqn. in question, one obtains:

D = 7.681i

x1 = 0.578 + 1.001i

x2 = 1.73 – 2.997i

and

a1 = 1.103 - 0.665i

a2 = 1.103 + 0.665i

a3 = -1.206

IF the Roots are correct THEN:

a1 + a2 + a3 = 1

and:

(a1 x a2) + (a 2 x a3) + (a3 x a1) = q

AND

a1 x a2 x a3 = r

AND:

(a1 – a2) x (a2 – a3)x (a3 – a1) = D

We will just check the first: a1 + a2 + a3 = 1

We have then:

[1.103 - 0.665i] + [1.103 + 0.665i] + (-1.206) = 2.206 - 1.206 = 1

Energetic readers can check the others!

Energetic readers can check the others!

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