The Project Gutenberg eBook #36640: Lectures on Elementary Mathematics

a³ − b³
+ ₂ (


³ a²b − ab²) = ^{1 √}−27q² − 4p³

can be made a perfect cube by adding the left-hand side of the equation

ab(a + b) = q,

multiplied by ³√⁻³ , and that the root of this cube is

2

1 − ^√−3 _{b −} 1 + ^√−3 _a

2

2

so that, extracting the cube root of both sides, we shall have the

expression
√

√

1

− −3 _{b −} 1 + −3 _a

expressed in known quantities. And since the radical ^√−3 may
2

2

also be taken negatively, we shall also have the expression

1 + ^√−3 _{b −} 1 − ^√−3 _a
2

2

expressed in known quantities, from which the values of a and b

can be deduced. And these values will contain the imaginary quantity ^√−3, which was introduced by multiplication, and will be reducible to the same form with the two roots

m^√3 A + n^√3 B and n^√3 A + m^√3 B,

which we found above. The third root

c = −a − b

will then be expressed by ^√3 A + ^√3 B.

By this method we see that the imaginary quantities em- ployed have simply served to facilitate the extraction of the
cube root without which we could not determine separately the values of a and b. And since it is apparently impossible to at- tain this object by a different method, we may regard it as a demonstrated truth that the general expression of the roots of an equation of the third degree in the irreducible case cannot be rendered independent of imaginary quantities.

Let us now pass to equations of the fourth degree. We have already said that the artifice which was originally employed for resolving these equations consisted in so arranging them that the square root of the two sides could be extracted, by which they were reduced to equations of the second degree. The following is the procedure employed. Let

x⁴ + px² + qx + r = 0

be the general equation of the fourth degree deprived of its sec- ond term, which can always be eliminated, as you know, by increasing or diminishing the roots by a suitable quantity. Let the equation be put in the form

x⁴ = −px² − qx − r,

and to each side let there be added the terms 2x²y + y², which contain a new undetermined quantity y but which still leave the left-hand side of the equation a square. We shall then have

(x² + y)² = (2y − p)x² − qx + y² − r.

We must now make the right-hand side also a square. To this end it is necessary that

4(2y − p)(y² − r) = q²,
in which case the square root of the right-hand side will have the form