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a being a root of the equation y1-1=0. Therefore, observing that a11=1 and r"=1,

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u。, u1, &c., being rational and integral functions of r which do not change by the substitution of ra, ra2, ra3, &c., in the place of r; for these quantities, regarded as functions of x1, x2, 3, &c., do not alter by the simultaneous changes of x, into x2, x1⁄2 into r ̧, &c., nor by the simultaneous changes of x1 into x ̧, into 24, &c., to which correspond the changes of r into ra, into ra2, &c. Now every rational and integral function of r, in which r"=1 may be reJuced to the form

I 2

1

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the coefficients A, B, C,... N being given quantities independent of r; or, since in this case the powers r, r2, r3, ...72-1 may be represented, although in a different order, by r, ra, ra2,... ra1-2, we may reduce every rational function of r to the form

A+Br+Cra+Dra2+... +Nrans

Therefore, if this function is such that it remains unaltered when r is changed into ra, it follows that the new form

A+Bra+Cras+Dra3+...+Nr,

coincides with the preceding;

... B=C, C=D, D=E, &c., N=B,

and therefore the function is reduced to the form

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since the sum of the roots =— -1; hence each of the quantities uo, U1, U2, &c., will be of the form A-B, and its value will be found by the actual development of z=y"; so that we have the case where the values of uo, U1, U2, &c., are known immediately, without depending upon the solution of any equation. Hence, if we denote by 1, a, ẞ, y, &c., the n-1 roots of the equation Ãa1—1—0, and by Z0, Z1, Z2, &c., the value of z answering to the substitution of these roots in the place of a in equation (1), we shall have, as in the former cases,

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an expression for one of the roots of the equation x-1=0; and the other roots are r2, 73, &c.

Thus, the solution of x-1=0 is reduced to that of the inferior equation y"--1=0, of which 1, a, ß, y, &c., are the roots; also, since n-1 is a composite number, the determination of a, ß, y, &c., will not require the solution of an equation of a higher degree than the greatest prime number in n-1; that is, the solution of TMa—1=0 (n prime) may be made to depend upon the solution of equations whose degrees do not exceed the greatest prime number, which is a divisor of n-1.

EXAMPLE IV.
x-1=0.

The least primitive root of 5 is 2; for the powers of 2 from 1 to 4, when divided by 5, leave remainders 2, 4, 3, 1;

..y=r+ar2+a2r1+a3μ3;

a1=1, r3=1, and r+r2+r++r3=−1;

2y=-1+4a+14a2-16a3.

also

But the four roots of y1-1=0 are

1,-1,-1,-√-1;

..zo=1, z=25, z=-15+20√-1,
23-15-20 √-1;

..x=1{−1+√5+ √15+20+15-201}.

375. For the proof that, in the general equation of the nth degree, the formation of the reducing equation will require the solution of an equation of

1.2.3...N

(m-1)m(1.2.3...p

1.2.3...(n—2) dimensions, when ʼn is prime; and of dimensions, when n is a composite number, and =mp, where m is prime; and that, consequently, the method fails when n exceeds 4, the reader is referred to Lagrange's Traité de la resolution des équations numériques, note xiii., from which the matter of this section is taken.

RESOLUTION OF THE GENERAL EQUATIONS OF THE THIRD AND FOURTH DEGREES.

RESOLUTION OF THE EQUATION OF THE THIRD DEGREE.

376. I shall suppose that we have made the second term of the equation of the third degree disappear, and, to avoid fractions, I will write this equation under the form

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Among the different modes of resolving it, the most simple consists in forming a priori an equation of the third degree, without a second term, which admits of one known root, but expressed with indeterminates, and to make use afterward of these indeterminates to render the equation identical with the proposed equation (1). To establish this identity, it will be necessary to write two equalities, and for this reason we employ two indeterminates. Let there be made x=a+b: the cube will be 3=a3+b3+3ab(a+b); then, replacing a+b by x, and transposing, we shall have

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an equation which admits the root x=a+b, and which it is necessary to render identical with equation (1). Therefore we place

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The first of these equalities gives a3b3=-p3. Thus we know the sum a+b3, and the product ab3. Then the values of a3 and b3 are roots of an equation of the second degree, in which the coefficient of the second term is equal to +2q, and the last term equal to -p3 (see Art. 191); so that this equation will be, calling z the unknown,

222qz-p=0.

This is called the reduced equation.

Its two roots represent the values of a3 and b3; moreover, we can take either of them indifferently for the value of a3, because this amounts to changing a into b, and b into a, in the value x=a+b. I will take

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Each radical of the second degree here has but one value, but each one of the third degree has three. If we could satisfy equation (3) without making any choice between these values, we could also, by the same values, render equation (1) identical with equation (2); and since a+b is a root of the second, the first ought to be satisfied by taking

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But an important remark presents itself: it is, that since each radical of the third degree has three values, the above expression must have nine, while the equation (1) ought to have but three roots. It is necessary to explain, then, whence comes this multiplicity of values, and to discern among them which ought to be true roots of the equation (1).

For this purpose, let us observe that, properly speaking, it is not the resolution of equations (3) which has given a and b, but rather the equations

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Now if we designate by a and a2 the two imaginary cubic roots of unity, which, as we know, are the one the square of the other, it will be readily seen that the equation a3b3=—p3 may result indifferently, from raising to the cube these following:

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Hence it follows that the nine values contained in formula (4) ought to give the roots of the three equations,

x+3px+2q=0, x+3apx+2q=0, x+3a2pr+2q=0.・・・ (6) We can, moreover, consider these nine values as the roots of the equation of the 9° degree, which would be obtained by multiplying together the three equations (6). But it will be more simple, and will amount to the same thing, to raise to the cube either one of these equations, after transposing to the second member the term which contains p. In this manner we find at once

(x+2)=-27p3x3.

As to the roots which belong especially to each of the three equations, what precedes furnishes the means of distinguishing them; because, according as the coefficient of x shall be 3p, 3ap, or 3a2p, it is clear that we ought to add only the values of a and b, for which we have ab=-p, or ab=-ap, or ab=a2p.

By this rule it will be easy to form the roots of the proposed equation x3+3px+2q=0, the only one with which we have to do. Designate by A one of the values of the first cubic radical, and by B one of the values of the second; the values of a and b will be

a=A, aA, a2A; b=B, aB, a2B.

Moreover, suppose, for this is admissible, that A and B represent the values, the product of which is -p. From what has just been said we ought to add only the values, the product of which is AB; then, recollecting that a3=1, we must take

x=A+B, x=aA+a2B, x=aaA+aB;

and, besides, we know (303) that we have

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If we replace A and B by the two cubic radicals, and a and a3 by their values, we shall have

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These are the roots of the proposed equation, but we must take care to attach to the two cubic radicals the same restricted sense as to A and B, without which we should find false roots.

377. To discuss these values, it will be more convenient to leave A and B substituted for the cubic radicals, and to isolate the one which is multiplied by √3. By this means we have

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I shall suppose, also, as is done ordinarily, that the coefficients 3p and 24 represent real quantities. Then equation (1), being of an uneven degree, has always one real root, and it is admissible to suppose that A and B are the values of a and b, which give this root; so that A+B will be a real quantity. This being premised, let us return to the two radicals

3

a=V−q+ √ q2+p3, b=√ −q−√ q2+p3.

If q2+p3>0, each of them has one real value; then we can suppose A and B real. Consequently, A+B and A-B will be so also; then the first root x=A+B is real, and the other two are imaginary.

If q2+p3=0, we have A=B, and then the three roots will be x=2A, x=—A, x=—A. They are all three real, and the last two are equal with one another.

Finally, let q+p3<0, which requires p to be negative. Then a and b have no longer any real determination, and, consequently, the three values of x are found complicated with imaginary quantities. However, we know that one of them must be real, and, indeed, it is evident that the cases in which the three roots of equation (1) are real and unequal can only be found on the hypothesis in question, that q2+p3<0, as may be seen by referring to the supposition just above of q2+p3>0. It would be wrong, then, to affirm that the values of r are imaginary. I will prove, in fact, that neither of them are so; and as we can always suppose that A and B are determinations such that the sum A+B represents the real root, the existence of which is demonstrated, the whole is reduced to showing that the part (A−B) √ —3, which

is found in the other two values of x, must be real. By the rules of algebra alone we have (A—B)(A2+AB+B)=A3-B3; then

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But, because of the values of a3 and of b3, we have A3—B3=2 √ qo+p3; and, by the manner in which A and B have been chosen, we have AB=—p;

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But by hypothesis we have q2+p3 <0; then the quantity above is real; then the three values of r are also.

It is thus demonstrated that, upon the hypothesis of q2+p3<0, the imaginary quantities which affect the three values of r must destroy one another. It would seem, therefore, that analysis ought to furnish the means of making them disappear, but as yet it has not been found capable of effecting this reduction. For this reason, the case under examination has been called the irreducible case. Whenever the equation falls under this case, the general expressions of the roots will be of no use in calculating their numerical values, and then we can recur to the methods of Arts. 290–297.

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7

9

We have p――2, q=— · · √q°+p2 = 2; which gives

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r=−1+√3 √ −1+¦(−1−√3 √√−1)=!(−3+ √3 √ −1),
r=—1—√3 √ −1+1⁄2(−1+√3√ −1)=1(—3—√3 √ −1).

(2) Here

x3-21x+20=0.
p=-7, q=10;

3

-10+9√3+V-10-9√3.

This example is one of the irreducible case. The general value of x appears in an imaginary form, and yet the roots are real, being the numbers 1, 4, and -5, which, by substitution, will be found to verify the given equation. 378. The solution of the irreducible case may be obtained, also, by the help of a table of sines and cosines. We subjoin the method, for the benefit of the student acquainted with trigonometry.

Solution of the irreducible case by trigonometry.

cos 20-2 cos2 0-1

cos 30 2 cos 20 cos 0- cos 0

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