equation, and we have therefore z=3, y=9, and x=36. ..21296, the greater number,

and 720=576, the lesser.

The other two values of r are found by trial to be impossible.

EXAMPLES FOR PRACTICE.

1. Find the three values of y in the equation y3=1.

2. Given x3-36x-91=0, to find the values of x.

3. Given x3-28x+48=0, to find the values of x.

4. Given +10x2+31x+30=0, to find the values of a.

ANS. x2, or -3, or-5.

5. Given y3+24y=245, to find the values of y.

6. Given a3+18 x2 + 216 x − 33920, to find the values

of x.

ANS. x8, or −13± √−255.

2x3 412x

7. Given

+

10a2+224, to find the values of x.

ANS. x7, or 8, or 10.

8. Given a3-3x-18=0, to find the values of x.

9. Given x3-27x+54-0, to find the values of x.

440. WHEN all the fractions have been removed from an equation of the third degree, in the manner explained in the preceding chapter, and none of the divisors of the last term are found to be a root of the equation, it is a certain proof, not only that the equation has no root in integer numbers, but also that a fractional root cannot exist, which may be proved thus:

Let there be given the equation x3-ax2+bx—c=0, in which a, b, c, express integer numbers. If we suppose, for

first only has 8 for the denominator, the other being either integer numbers, or numbers divided only by 4 or by 2, and therefore cannot make 0 with the first term; and the same will happen with every other fraction.

(*) This is a case in which the number 3 is twice a root of the equation; the quantity 3-27x+54 being composed of the factors (x−3) × (x−3) x(x+6).

As

As in those fractions the roots of the equation are neither integer numbers, nor fractions, they are irrational, and, as it often happens, imaginary. The manner, therefore, of expressing them, and of determining the radical signs which affect them, forms a very important point, and deserves to be carefully explained. This method, having been first discovered by Scipio Ferreo for certain forms of cubics, was afterwards generalized by Cardan, and is usually termed Cardan's Rule.

441. In order to understand this rule fully, we must first attentively consider the nature of a cube, whose root is a binomial.

Let a+b be the root; then the cube of it will be a3 + 3a2b+3ab2+b3; and we see that it is composed of the cubes of the two terms of the binomial, and also, of the two middle terms, 3 ab+3ab2, which have the common factor 3ab, multiplying the other factor a+b; that is to say, the two terms contain thrice the product of the two terms of the binomial, multiplied by their sum.

Let us now suppose x=a+b; taking the cube of each side we have, a3 +b3+3ab (a+b): and since a+b=x, we have x3 a3+b+3abx, or x3-3abx+a+b, one of the roots of which we know to be x=a+b. Whenever, therefore, such an equation occurs, we may assign one of its roots.

For example, let a=2 and b=3, we shall then have the equation x3=18x+35, which we know with certainty to have x=5 for one of its roots.

Further, let us now suppose a3=p and b3-q, we shall then have a/p and bq; consequently ab=/pq; therefore, whenever we meet with an equation of the form of x3-3x2/pq+p+q, we know that one of the roots is p+q.

Now, we can determine Ρ and q in such a manner, that both 3pq and p+q may be quantities equal to determinate numbers;

numbers; so that we may always resolve an equation of this

nature.

442. Let in general the equation,

x3-fx+g.

be proposed. Here, then, it will be necessary to compare ƒ with 3/pq, and g with p+q; that is, we must determine p and q in such a manner that 3pq may become equal to ƒ, and p+q=g, for we then know that one of the roots of our equation will be 3/p+ V/q.

We have therefore to resolve these two equations;

4

There remains p2 - 2 p q + q2=g2—

g2.

Extracting the root, p-q=±√ 8° -27

27

443. In any cubic equation, therefore, of the form of fr+g, we have always for one of the roots,

that is an irrational quantity, containing not only the sign of the square root, but also the sign of the cube root; and this is the formula which is called the Rule of Cardan.

Let us apply it to some examples, in order that its use may be better understood.

Let x=6x+9.

Here we have ƒ=6 and g=9; so that g281 and ƒ3=216,

we have ƒ=3 and g=2, .. —,ƒ3=4 and g2=4, which gives

f3

4

27

27

f0; whence it follows that one of the roots is

444. To this rule it might, however, be objected, that it does not extend to all equations of the third degree, since the second term of the equation, or x2, is wanting. It remains, therefore,