4. By adding abc – abc to the dividend, it becomes by arrangement of the terms ab(a-b+c)+ac(a−b+c) + bc(a-b+c) which is obviously divisible by a- −b+c.

5. The dividend is (a✦ − b1)(a* +a2b2 +b1) and the divisor is a +c2b2+ba. 6. The quotient is a2 +b2+c2 - ab — ac – bc.

Deduce from this quotient, the three quotients (1) when a is written for + (2) -b for +b, and (3) -c for +c: and shew that the sum of the four quotients is 4(a2+b2+c2).

7. The dividend when reduced is 2(a3 +b3 +c3-3abc).

8. a2+b2+c2 + d2 — ab - ac-ad-bc-bd-cd.

9. By a like artifice as in 4, it is shewn the quotient is ab- bc + ac.

10. The quotient is (a−b)(a+b+c).

11. The first dividend involves the factor b-c which is common to the divisor, which may be removed, and the quotient is (ab+ac+bc).

The second dividend, when the terms are expanded, will be found to contain the same factor as one of the factors of the divisor. This common factor being removed, the quotient is 5{a2 − (b + c)a + (b2 − bc+c2)}.

1. (b+a)(b+c) = b2 + (a+c)b+ac

(d+c)(d+a)=d2 + (a+c)d+ac

•'. (b+a)(b+c)− (d+c)(d+a) = (b2 − d2) + (a+c)(b − d)

• '. (b+a)(b+c)− (d+c)(d+a) − (a+c) (b − d) = b2 — d2.

2. (a+b)2+2(a−b)(c−d) + (c+d)2. 3. (a+b+c+d)2 + 2(a2+b2+c2+d2).

4. 4(a2+b2+c2+d2). 5. (a−b)3. 6. 0. 7. a2-d2.

8. Bracket a+b and c+d, so as to make each factor a binomial. In two lines the aggregate is shewn to be ab+cd.

9. (ac+bd) (bc+ad). 10. (a*—b1)(c1 —d1). 11. (a2+b2)2.(c2+d2)2.

12. a+b+c4 + 4abcd.

13. The six binomials when expanded can be thus exhibited—

3a3+3ab(a+b)=3a(a2 + b2) +3a2b
3b3 +3bc(b+c)=3b(b2 +c2)+3b2c
3c3 +3cd(c+d)=3c(c2+d2)+3cd'
3d+3da(d+a)=3d(d2 + a2)+3d2a

3ac(a+c)+3bd(b+c)

Next if 3a2(a+c)+3bd(b+c) be combined with 3a2b+3bgc+3cd2 + 3d3α they become 3a(c2 + d2) + 3b(a2 + d2) +3c(b2 + α2) + 3d(b2+c2)

which when added to 32(a2+b2)+3b(b2+c2) +3c(c2 + d2) + 3d (d2+a3)

make up 3(a+b+c+d) (a2+b2+c2+d3).

14. (2a)3. 15. 6abc. 16. 24abc.

17. There are twelve terms in this expression to be combined. First combine the fifth with the first and the sixth with the eighth, and then the two results.

Next combine the third and the seventh and the fourth and the twelfth. Thirdly, reduce the results from the first and fifth and sixth and eighth, with that of the ninth, tenth, and eleventh. Then the results from the third and seventh, fourth and twelfth, and the second.

These results being one set positive and the other negative, when combined leave the remainder 24abcd.

18. (a-b)(a–c)(a–d)(b−c)(b−d)(c–d).

19. This reduction can be effected in the same manner as in 12.

The result is 60abcd(a+b+c+d).

XII.

The verifications of the expressions under this number may be readily effected by performing the indicated operations.

XIII.

The exercises under this number will occasion no difficulty. The identical expressions may be expanded or resolved into the same factors.

9. The three expressions under this number are each reducible to

a® +be+c® − 6(a*bc+b*ac+c1ab) + 2(a3b3 + a3ç3 +b3c3)+9a2b3c2.

XIV.

The exercises under number require no directions further than to employ any methods whereby the labour of reducing the expressions may be lessened.

XV.

1. Make the proposed substitutions, which offer no difficulty.

2. Then A+B+C=a2+b2+c2-ab-ac-bc

and B-AC=b(b3 +c3+a3 — 3abc)=b(a+b+c) (a2 + b2 + c2 — ab-ac-bc)

A-BC and

C2-AB

are equal to the

= (a+b+c)(A+B+C)

C

and (B2 — AC)ac = ( A2 – BC)bc=(C2 — AB)ab=abc(a+b+c)(A+B+C).

3. Here A3+2B3 — 3 AB2 = (A−B)2(A+2B)

= (a2 + b2 + c2 — ab−ac—bc)2(a+b+c)2.

4. In order that AB(A2 + B2) may be equal to CD(C2+D2), (a+b)* − (c+d)* must be equal to (ab)—(c-d), whence it may be shewn that ab(a2+b2) = cd(c2 +d2).

2. b2+c2+2bc - a2 − (b + c)2 − a2 = (b+c+a)(b+c-a); substitute for a, b, c. 5. By adding the three equalities, (x − y)(y − z)(z − x) = a3 + b3 + c3

and (x − y)(y − z)(x-2)=3abc. Therefore a3 +b3 + c3 − 3abc=0.

6. a1+b2+c✦ −2b1c2 – 2a2b2 — 2a2c2 = - {2a2b2+2b2c2 + 2a2c2 — a1 — ba — c^} = − (a + b + c)(a+b−c)(a+c−b)(b+c− a)

=

And 2a +26+2c=2x+2y+2z, and a+b+c=x+y+z,

also a+c-b=y, a+b¬c=z, b+c-a=x.

Hence the expression is equal to -(x+y+z)xyz in terms of x, y, z ;

and (x+y+z)(xy + xz+yz) − xyz=(x+y)(x+z) (y + z) = 2a. 2b. 2c=8abc.

7. From the equality of the first and second terms, −m=x2+xy+y2, and from

the equality of the second and third, -m=z2+yz+y2.

Hence z2+zy + y2=y2+xy+x2, and (≈2 − x2)+y(z−x)=0.•.x+y+z=0.

• ' •y3 + z3 +m(y + z) = (y + z) { y2 − yz + z2+m} = (y + z) (y2 − yz +z2 − z2 − yz− y3)

=(y + z)( - 2yz)

=(-x)( -2yz) = +2xyz.

The same value may be found by a similar process for the other two expressions. If mxy, &c., be substituted for m(x+y), &c., in this exercise, so that the equivalents become y3+z3+myz=z3 +x3 +mzx=x3 +y3+mxy, it may be shewn that m=x+y+z, and x2 + y2 +z2 + yz+xz+xy=0, also that each member of these equations is equal to xyz − (x+y)(y+z)(z+x).

8. s(s-a)(s-b)(s − c)=2a2b2 +262c2+2a2c2 -a4-b4-c4. Make the substitutions for a2, b2, c2.

11. x2 + y2=p2 — 2q, x3+y3=p3 −3pq, x2+y* =p4 − 4p2q+2q2,•

x+y=p5-5p3q+5pq2, &c.

12. Multiply the factors and then make the substitution.

13. Multiply the first, second, and third equalities by x, y, z respectively, and add the results.

1. z.

2.

1- -x 1-2x

XVIII.

which when the division is performed will give the series.

Square each expression and take their sum.

3. x2y=z(x−z + y)2 = z(x − 2)2 + 2yz(x − z) +zy3

• • . z(x − z)2 = y(x2 − 2xz+z2)+yz(z − y),

•' . ( z − y) (x − ≈)2=yz(z− y); .'. (x − z)2 = yz.

4. Since x+y+z−xyz=2, .'.x2 + xy +xz− x2yz=2x,

.* . - 2x+x2 = −xy − xz + x2yz,

add unity to each, .' . 1 − 2x + x2 = 1 − xy−xz+x2yz.

5. Since b(bx2+a2y)=a(ay2+ab2x), then b2x2 - a2y2 =ab2x − a2ly,

or (bx+ay) (bx − ay)=ab(bx – ay),

•*. (bx — ay) { bx+ay — ab} ==0

... bx-ay=0 and bx=ay

also bx+ay - ab=0 and bx+ay=ab.

6. (a+b)x−(c+d)x=cd-ab; and (a+b)x+ab=(c+d)x+cd, add x2 to each of these equals, and x2 + (a+b)x+ab=x2+(c+d)x+cd.

7. The required expression may be thus arranged—

ay(ax+by − 1)+ bx(ax+by − 1) + (ax+by − 1) =0

or (ay + bx+1)(ax+by-1)=0.

8. x= − (y + z), .'. − (a2 − bc)(y + z) + (b2 − ca)y +(c2 −ab)z=0

and (b2 - ca - a2 + bc)y = (a2 − be — c2 + ab)z

or (b− a)(b+a+c)y=(a−c)(a+c+b)z.·. (b − a)y=(a–c)z or by-ay-az-cz ...by+c=a(y + z) = − ax... ax+by+cz=0.

9. Expand, and remove common quantities, then arrange the terms, and the expression is divisible by a2 - b2.

11. From a+b+c=0, may be found (a2+b2+c2) = − (a2+ab+b2),

}(a3 +b3 +b3) = − abc and †(a3 +b3+c3)=abc(a2+ab+b2).

12. From the first two equations a(x+y)= b2, and x+y=c, .'.ac = b2.

13. Subtract the second expression from the first and divide the difference by y-z, and then a=x+y+z.

14. From the first and second members of the given equality, by − (a−b)z=a*, from the first and third, cz – (a–c)y=ax, and from the second and third,

- (b−c)x=by; these will give ba⇒ay and bz=cy, also by=by;

Cz

... b(x+y+z)=(a+b+c)y=0.

15. b2y2 - 2bcxy+c2x2 = b2y2 − b2cz — acy2 + ac2z2.

Then - 2bcxy+c2x2 = b2cz - acy2+ac222, by removing by.
-2bxy+cx2 = −b2z-ay2+acz2 dividing by c.

-2abxy+acx2 = − b2az+a2 y2+a2cz, multiplying by a.

.*. b2x2 – Qalxy+a3y2=b2x2 − acx2 − ab2z+a2cz, adding b2x2, and (bx-ay)=(b2 — ac)(x2 — cz).

17. From the given equality may be deduced,

(x2 + y2+z2)(xyz+1)=(xyz+1)(xyz+4).

In a similar way a” – b" may be shewn to be greater than nb"-1(a - b).

3. Here ama” = a”(am−n − 1) supposing m>n.

And am – 1 is divisible by a +1 when m-n is even.

Similarly for the rest.

5. The product mnp has the divisors m, n, p, mn, np, mp, and m, n, p may be in each case odd or even numbers.

· 7. x2 - na2-1x+ (n − 1)a′′ = (x′′ − a*) — na®−1(x − a)

= (x − a) { xn−1 +xn−2ɑ+x2-3a2+.

+a^-1 — na"−1}

= (x − a) { (xn− 1 − a®−1) + (xn−3 — an−2)a + (xn−3 − a*−3)a2+ . . . .

The quotient is x3 + 2x3a + 2xa3 +3a3.

to n terms}.

-

8. If n1 be an odd number, then a-1-b-1 is divisible by a-b, and an – ba is divisible both by a-b, and a+b; and it is to be shewn that a+1 — ba+1 is divisible by (a−b)(a2+ab+b2).

XX.

1. It is obvious that bx+cy must be one of the factors, and that x2+axy+y2 must be divisible by bx+cy. After performing the division the remainder will be y2, which must be equal to zero, in order that the division may

{1-(ab-c)} y2

leave no remainder.

2. The coefficient of x4 is 3a4.

3. Four factors of the form x+a and one of the form x-b. See Section IV., Art. 12, pp. 19, 20, and the note.

4. x3+px2+px+1=(x3+1)+px(x+1); and x2+1 is always divisible by x+1 when n is an odd number.

5. See Section IV., Art. 13, p. 24.

XXI.

1. The first expression may be shewn greater or less than the second, third, and fourth according as bc is greater or less than a2, ac than b2, and ab than c2.

2. a= − (b+c), y=−(x+z), z=−(x+y) ; .'.a2yz=+ (b+c)2 (x+z)(x+y) and similarly, for b2xz and c2xy.

3. Suppose each of the three expressions to be positive; then the sum will be positive, and the terms may be arranged thus: (a−c)y + (c − b)x+(b− a)z, of which two of the terms are negative and one positive. But if the terms be arranged in this order: (y-z)a + (z − x)b + (xy)c, two of the terms are positive and one negative. Hence the hypothesis of all the three terms being positive is not sustained. 4. If ab, then a-b>0, and a2 – 2ab+b2 > 0. • .a2+ b2 > 2ab.

Next a2 - 2ab+b2>0, add 4ab to these unequals, and a2+2ab+b2> 4ab or (a+b)2 > 4ab.

5. a2+b2>2ab, a2+c2 >2ac, b2+c2>2bc.·.2(a2 +b2 +c2)>2(ab+ac+bc) or a2+b2+c2>ab+ ac+bc.

6. The aggregate of the three expressions on the left is equal to 3(a2+b2+c2)-2(ab+ ac+bc).

7. The polynomial arises from the product of the three factors (a+b−c)(a+c-b)(b+c− a), and is a positive quantity.. a+b>c, &c. 8. Since a >b, a2 + b2 > 2ab, and a2 + b2 > (a2 – ab+b2),

.*.(a2+b2)2 > 2ab(a2 — ab+b2).!

9. a2+b2 > 2ab, b2 +c2 > 2bc, a2 + c2 > 2ac,

•'. (a2+b2)(b2+c2) (c2 + a2) > 8a2b2c2, for a2, b2, c2, write a, b, c, and (a+b)(b+c)(c+a) > 8abc.

See XIII., 4. (a+b+c)(ab+bc+ac) − abc=(a+b)(b+c)(a+c),

and (a+b)(b+c)(c+a) may be shewn greater than 8abc,

.. (a+b+c)(ab+ac+bc) > 9abc.

10. Let ab and a >c, then a-b> 0 and a-c> 0, &c., ... (a - b) (a - c) > 0 or

a2 − ac - ba+bc > 0, and ... bc> ac+ab - a2 or bc>a(c+b-a).

Similarly ac>b(a+c− b) and ab>c(a+b−c).

Hence a2b2c2abc(c+b− a)(a+c-b)(a+b−c)

and... abc> (c+b− a)(a+c−b) (a+b−c).

This result illustrates the theorem: "The rectangular parallelopiped formed by the three sides, a, b, c of a triangle for its edges is greater in volume than that formed by the three differences between the sum of every two sides and the third side."

12. Except when a=b.

14. From the given expression (bx − ay)2>0; if bx> ay then

(x2+ y2)(a2+b2)>(ax+by)2.

15. (x+a)2 (x2+b2)=(x2 +a2)(x2 +b2)+2ax(x2 +b2),

and (x+b)2 (x2 +a2)=(x2 +b2)(x2 +a2)+2bx(x2+a2). The answer depends on whether 2ax(x2+b2) is greater or less than 2bx(x2+a2). If x2 >ab,

then (x+a)2(x2 +b2) is greater than (x+b)2 (x2 +a2).

16. x2y2 = (a2+b2)(c2 +d2) = (ac + bd)2 + (ad - bec)2, but if ad= bc,

then x2y2=(ac+bd)2 and xy=ac+bd.

17. If (a2 + b2 + c2)(x2 + y2 + z2) be greater than (ax + by + cz)2,

then (ay - bx)2 + (az − cz)2 + (bz − cy)2 is greater than 0, but the squares both of positive and negative quantities are positive, therefore each of these squares is positive, and consequently the first expression is greater than the second.

18. (a+b+c)3 a3 + b3 + c3 + 3(a+b) (b + c) (c + a); first shew a3 + b3 + c3 > 3abc, next 3(a + b) (b+c) (c + a) > 24abc from Ex. 9, XXI.

19. (1) (x+y-≈) 2 + (x + z − y)2 + (y + z − x)2 = 3(x2 + y2 + z2) − 2(xy + xz+yz), and x2 + y2 + z2 has been shewn greater than xy + xz + yz. (2) and (3) may in like manner be shewn to be true.

20. (1) Let x > y and y>z, then x2(y-z) and z2(x − y) are positive, and y2 (z – x) is negative; whence x2 (y − z) + z2 (x − y) > y2 (z − x) from which may be deduced, that x+zy, but x>y, therefore the property is proved. (2) and (3) may be considered in the same manner.