In the proposed cubic equation, which we may write under the form

put the unknown r cos for r; or, which is the same thing, put

for cos 0,

and

3

1

cos 30=-2q,

r2=—3p .. r=2 √ —p, which is real, p being negative;

Consequently, the trigonometrical solution of the proposed cubic equation, that is, the determination of 0, and thence of r cos 0, depends upon the trisection of an arc, or the determination of cos 0 from cos 30.

The mode of proceeding by aid of trigonometrical tables is obvious; we are

to seek in the table of cosines for the angle whose cosine is

this will

belongs equally to three arcs,

be the angle 30, and, consequently, one third of it will be ; and the cosine of this, multiplied by r, or 2 √—p, will give r cos 0=x for one of the real roots of equation (2). As the given cosine, viz., 30, 230, and 2π-30, by taking the cosine of one third of each of the latter two, we shall have the values of the remaining roots. Thus all the three roots will be expressed as follows:

-

-p3

1

2 √ −p cos 0, 2 √ −р cos (2π+30), 2 √−p cos (2π—30).

Or, using the supplements of the two latter arcs instead of the arcs themselves, and remembering that the cosine of an arc is equal to minus the cosine of its supplement, we have somewhat more simply the three values of x in the following form:

2√−p cos 0, −2 √ —p cos (60° —0), −2 √ —p cos (60°+0).

This method, with a single exception, applies to the irreducible case; for, as the trigonometrical cosine of an arc is always less than unity, except when that arc is a multiple of 180°, we must have

or

√

<1 ... q2 <—p3, q2+p3<0.

When 30 is a multiple of 180°, two roots must be equal.

The reducible case may also employ the aid of trigonometry.

WOOLLEY'S METHOD OF RESOLVING THE CUBIC EQUATION. 380. The following method of arriving at a new and valuable formula for the solution of cubic equations will be found an excellent exercise for the student :* Let the given equation be

*It is the production of an old pupil of the author's, Mr. James S. Woolley, whom ill health, and other discouraging circumstances, have not prevented from making some important discoveries in algebra, which it would be premature at present to publish to the world.

FF

The value of m, which renders the coefficient of w zero, may be found thus

3pm2+9qm—p2=0.

The value of w in (7), substituting the value of m, found in (8), is expressed in the following four equations, (9), (9, a), (9, b), (9, c), the last three being obtained by decomposing (9) into factors.

Substituting in (6) the values of m and w, found in (8) and (9, c), we shall

Substituting in (4) the values of z, given in (10), and decomposing one more of its terms into factors, we shall have

54

干

27

2

3

F

8192 12p+

t

2

27

4

F

+ » √ = p+ ( − 3 + √ √ = x+3)+ √12p+813(¦± √÷p+?)}

(continuing the numerator) (812+12) (→±√5+)

27 4

+

(12)

But the first term in the numerator of (12) may be transformed thus:

But the last term in the numerator of equation (12) is

(812+12) (→ 1 + √5+3).

Therefore the sum of the first and last terms of the numerator of (12) is

54

p2

we have

9

(-1+ √er+D)2

(−1 + √ √r + D) + ( − 1 + √ir + 9)'

The numerator of this value of x is equal to

Dividing numerator and denominator by the common factor, we have

3 P
3

(+)

This formula may be reduced to that of Cardan by dividing the numerator by the denominator, and observing that

But the first form is preferable, as it gives only the three values which satisfy

equation (1), whereas Cardan's formula gives nine values, six of which have

which is an advantageous form, inasmuch as but one third root has to be ex

tracted, both radicals having the same form.

A shorter solution of the above might be given, but we have already extended our article on cubics sufficiently far.

IRRATIONAL EXPRESSIONS ANALOGOUS TO THOSE OBTAINED IN THE RESOLUTION OF EQUATIONS OF THE THIRD DEGREE.

n

381. One of these expressions is VA± √B; but it frequently happens that A and B are rational numbers, and then it may be possible to reduce these radicals to simpler expressions, in which there are no longer radicals over radicals. This problem has already been resolved for radicals of the second degree, and it is now proposed to resolve it with reference to radicals of the higher degrees.

I shall commence with the cubic radical

A+VB. We can not suppose

for this root a quantity of the form √a+ √b, for we have

(√a+√b)3=a√a+зa √b+3b√a+b√b

=(a+3b) √a+(3a+b) √b,

a result which contains the radicals va and √b. But the preceding calculation shows that we should have a result of the form A+ √B, by raising to the third power the expression a+ √ and (a+ √b) √c. I will choose this last expression as the more general; we shall then have

√A+

A+ √B=(a+ √õ) Vē

(1)

Raising both members to the third power, it becomes A+ √B=c(a3+3ab) +c(3a2+b)√b; equating the rational parts together, and the irrational parts by themselves,

The problem, then, is, to find for a, b, c rational values which satisfy these two equations. But squaring these equations, and then subtracting the one from the other, we have

A-B-c2(a-3ab+3a2b2-b3)=c2 (a3—b)3 ;

Since a and b ought to be rational, it will be necessary to take e such that (A2-B)c be an entire or fractional cube, which is always possible. Calling the second member of the above equation M, we shall have a2-b=M, whence b-a-M. By substituting this value of b in equation (2), it will become