Ask anyone who took freshman
math, and they’ll tell you of their introduction to algebra - which is one of
the most useful branches of mathematics, since it paves the way for properly
doing derivatives, and integrations in calculus. And almost everyone taking it
would have been exposed to quadratic equations of the form:

ax

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

x = [-b

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

x

x

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

With:

x1 = [p

x2 = (p

The ROOTS are then – in turn:

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

WHERE:

r = - 1/2 + (-3)

r2 = -1/2 – (-3)

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

(and let interested readers check the others):

We have then:

ax

^{2}+ bx + c = 0which 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}]/ 2aBut 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 = 0What 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)/ x1The 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}Dr2 = -1/2 – (-3)

^{1/2}/2Last 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

(and let interested readers check the others):

We have then:

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

All of the above will be found to check out for the given equation!

All of the above will be found to check out for the given equation!

__PROBLEMS:__
1. For the example problem above, check that the roots
obtained satisfy each of the following:

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

AND

b) a1 x a2 x a3 = r

AND

b) a1 x a2 x a3 = r

2. Using the procedure formulated in the article, solve the
cubic equation:

x

^{3}– 6x^{2 }+ 11x – 6
Again, check to ensure the roots you obtained are the
correct ones!

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