XIII.
The operations under this number involve no difficulties.

XIV. The multiplications and divisions contain no anomalies which require remark.

XV.

The exercises under this number require no remarks.

+

=0.

ac

XVI. The first eleven examples nnder this number offer no difficulties. 12. In this example it may be shewn that. 2 2

2

+ (6 – c)(c - a)' (c-a)(a - b)(a - b)(6-c) 17. Each fraction may be simplified as the first, (a – b)2 – (6 – c)2 – (a – 6+6 – c)(a – -b+c)_ (a – c)(a + c - 26) _a+c - 26 a2 + ab - bc - c? (a? -- ca)+ b(a – c) (a – c)(a +c+b) a+b+c'

18. The shortest method here is to reduce each expression to the same identical form. Thus a? + 62 - C? 62 +c2 - a? c? +a? - 82 +2+ +2+

+2
ab

bc
(a + b)2 – (2_16+c)2 – a? (c+a)2 – 12

+
ab

bc (a+b+c)(a + b –c)+(6+c+a)(b+c-a)_ (c+a + b)(c+a-0) = (a+b+c). {a+b=c+b+c-a +c+a-b

ab

bc
(a+b+c)(2ac + 2bc + 2ab – a? – 62—(2)

abc
1 1 ?

+

ab'bc
4ab- (a + b)2 +2(a+b)e-c2

c++
(2ac + 2ab + 2bc-a? – 62 - c?)(a+b+c)

abc

+

ас

b

+

ab

ac

ac

Next {tab– (a+b+c)2}. (attēc + ca}
- {(*26*} {*****

.

XVII.

1. 0. 2. 0. 3. 0. 4. 0. 5. -1. 6. i. 7. +3. 8. a+b+c. 9. O.

2c 10. 0. 11.

12. 0. 13. 1. 14. a? +62 +c? + ac + ab + bc. (6 - c)(a - b) 15. a>(a+b)(a +c)_asla? +(6+c)a+bc),

21b+c)a

1+ (a - b)(a -c) a? -(b+c) + bc Similarly b8(6+c)(6+a)

=631+

2(a + c)6 (6 -c)(6-a)

(6 - c)(6

and

=C3

a-C

{ {

") },

c*(c + a)(c+b)

2(a + b)c?

1

(c- a)(c-b) Then

2(b + c)a“ 2(a + c)64 2 $(6+c)a4 _(a + c)b4 } (a−b)(a-c) (a - b)(b -c) a- a bl

b-C
2 S (09-c?)a+ - (a? -c)

(a -- c)(b - c)
2 aob(a? – 62) - co(a* - 64)
a-6°1

(a - c)(b -c)
2{a2b2(a+b) – c* (a? +62)(a +b)},

(a - c)(6-c) 2{a2b2(a+b) – c?(912 +62)(a + b)} And

2(a + b)c+
(a - c)(b-c)

(c-a)(c-b)
2{(a+b)e« – (a + b)(a +62)(3+(+ b)^232),

(c-a)(c-6)
2(a+b){c^ – (a2 +62)ca +a%b2},

(c - a)(c-b)
2(a + b)(c? – Q2)(c? – 62)

(c - a)(C-6)

=2(a+b)(b + c)(c+a). And the sum of the three fractions is as + b3 + c3 + 2(a + b)(b+c)(c + a).

XVIII.

2)=c(1–0) }

1. 0. 2.
322 - (a? +62 +c?)

3. 0. 4. 1.
(x – a)(2-6)(x–c)
5. First

x-clx-h_x-a x-cs - x(a - b) + (a? — 12) – c(a - b)
a-bla-c b-cla-02

(a — c)(6–c)
-(a-c)(x-a-b+c)

(a - c)(b-c) Next

(x-a)(x – b)_ (x – c)(x-a-b+c)
(c-a)(c-6) (a - c)(6–c)
22 – (a+b)x+ab - {x2 – (a+b)x+(a+b)c –r?},

(a - c)(6 – c)
(? – (a + b)c + ab=1.

(a – c)(b -c) 6. 1. 7. 22. 8. (x – a)+(y – 6) + (z – c).

(52 – ac)y 10.

(ax – by) (by – cz)

XIX.

1
1
1

1
1.
2.
3.

4.
abc
abc (x + a)(a+b)(x + c) (x - a)(x - b)(x -- c)

(a+b+c) 5.

6. (a + x)(b + x)(c + x) (a + b)(6+c)(c + a) 7. Sometimes by separating each fraction into two parts, the process may be simplified. b+c-a

b+c (o + c)(c-a)(a - b) (6 + c)(c - a)(a-6) (6+c)(c - a)(a - b)

1 (c-a)(a - b) (6+c)(0-a)(a – 6)*

a

a

cta-6

1 Similarly

b
(c + a)(a - b)(6 -c) (a - b)(6-c) (c+ a)(a - b)(6-c)'

1
a+b-C
and
(a + b)(0-c)(c-a) (6-c)(c-a) (a + b)(6-cc-a)
1
1

1 Then,

=0. Sec. XVII. (c-a)(a - b) *la-6)(6—c)*(6—c)(c-a) And the aggregate of

6

+

+

Es. I.

will be found to be

cla? + ab + b2 - c?)

(a + b)(62 – ca)(c2 – a?) 8. Add the first two fractions, next the other two, and the sum of the results is

zero,

4(a? – ca).

ac

XX. 1. 68 + (a + c)b? – (a + c)26—(a – c)?(a +c). (6+c-a)(a + b – c)(c+a-6) 8abc

1 2.

3.

4. 4.
(a? +62 +c2)2 - 4(a2b2 +a?c? + 62c2) C+a-6
2(a+b+c)abc

4(6-0) 5.

6.
7.

8. 1.
(6+c-a)(c+a - b)(a + b - c)* a +6-c
9.
(6+a - c)(a +c-6)(b+c-a)

(a+b+c)* 10. Let y=b+c-2a =(c-a)-(a-)

z=c+a - 2b = (a - b) – (6-c)

x=a+b- 2c= (b − c) — (c-c)
..x+y+z=0,

- Y
also c-a=x+y= 6-c=x+z-
2

2

2x a+b - 2c, b+c-2a , cta - 26 Hence

2y

22 +

+

+
y+2-20 2+- Y X+Y--Z
2x 2y

22
+

3.
- 2.0
- 2y

2
6-c

-- 2x

- 22
Similarly
+

+
+

-3.
a+b-2cb+- 2a ' cta-26 2x

2Hence the product of the quantities --3X-3= +9.

a-bey+2—2

2

+

a-6

b-0

C-a

+

a-6

C-a

+

2y

4.

XXI.
(a+b+d)(a + b)-cc+d) _ (a+b)? — c? +(a + b-c)d
(a+b+d)(a +d)-0(6+c) (a +d)? -c? + (a + d-c)b

(a+b+c)(a +b-c)+(a + b^c)d
(a +d+ c)(a + d -c)+(a +d-c)6
(a+b—c)(a+b+c+d) _ a+b-c

(u +d-c)(a +d+c+b) a-c+d 8. The fraction when reduced is

2abe(1

+ a2 + b2 +ca)

4(1 + a2 +62 +c?) 9. By means of the formula 23 – y: = (x— Y)(x2 + xy + y2), the expression is reduced to

9(6--c)(a? +62 +c? - ab-bc-ca)

(6+c-2a)(a? +62 +c? - ab-ac- bc) 10. By inspection it may be seen that b-c is a common factor of the numerator and denominator. When this factor is removed, the fraction is (b+c)as - (69 + bc+c?)+ b2c2

2

c

2

11. After reducing and arranging the quantities according to the descending powers of a, the highest common divisor will be found to be

a? -(b+c)a +(62 — bc+ca). 12. Here (ab+cd)(a? +62—(? – da) – (ac+bd)(a? – 82 +c? - da)

(a? + 62 - (? - d)(a? – 62 +c? - d2) + 4(ab+cd)(ac+bd)

(ac + bd +ac+bd)(62 - c2)+(ab+cd - ac- bd)(a? – dla) {(a?-da) + (62 -c)}. {(a?-da) - 162 - c')} + 4(a?bc + baul + c'ad+dbc)

(a + d)(b + c)(12 --ca) + (a - d)(b - c)(a? - da) (a?-da)2 – (62 - c2)2 + 4ad(62 +ca) + 4bc(a? + d2)

(a+d)(b – c){(6+c)(6+c)+(a -d)(a - d)} (a?-da)? – (62 - c?)? +(a +d)?(6+c)2 – (6-c)? (a - d):

(a + d)(6 – c). {(6+c)2 + (a - d)} {(a+d)? – (6–c)2},{(6+c)2 + (a – d)"}

(a + d)(b-c)
(a + d)2-(6-c)2
15. The first expression may be arranged in the form

(c?— 62) {a“ – (co +62)a? +Boc2} _(c+b)(c - b)(ao – bo)(a’ – c^).
(c-b){a: - (c+b)a + bc)

(a - b)(a -c)

=(c+b)(a + b)(a +c). The second in the same manner.

- 3(12 – c^)a:+3(64 –c)a? – 382c2(12 – c?)

- 3(b - c)a? + 3(62 – c2)a – 3bc6 –c)
(b + c)(at - (62+¢?)a: +bc9} _{0 + c)(a? – 12)(a:-) = (b + c)(a+b)(a+c).

{? (6+ ?
a: - (b+c)a + bc

(a−b)(a -c) 16. The given expression when reduced to an improper fraction becomes (1 + m + )(z - x)+ (1+ m? + x)(x - 6)^+ (1+ m + n”)== - {= + (c - a) + (a - b)}*

1+m? + n2
and when the terms of the numerator are expanded, they are

(1 + m2 + no)(x - a)?=(x - a)? + mo(x – a)2 + n2(x - a)?,
(1 + m + 2)(y−b)2=(-6)?+ m2(x - 6)^+ (y - b)?,

(1 + mo + n2)zo=22+ m2^2 + n2-2,
- {z+m(x − a)+n(y -b)}:=-22 – 2m2(x – a) – 2nz(y – 6) – m2(x – a)?

- 2mn(x - a)(y-b) – n°(y – )2. Here 32 - z, mo(x – a). – m?(x – a)', no(y – 6)? – no(y – 12) disappear, and the remaining nine terms may be arranged as follows :

(x - a)? – 2m2(x – a)+ mox?=(x – Q -- mz)",
(y-2)2—2nz(y—b) + n2z2=(y-b-nz)2,

no(x-1)2 – 2mn(x—a)(y—b) +mo(y—b)?= {n(x—a) – m(y—b)}":
Hence the given expression becomes by substitution
(x – mz—a)2+(y-nz—b)2+{n(x-a) – m(y—}”.

1+ m+ n2

XXII. 1. (r— 1)(. — 2)(x − 3)(x-4)

2.
s ** + a' ?

:}3. 0.
(x + 1)(x + 2)(x+3)(x + 4)

3* --* 4. 1+ 3 - 4x + x2 -- 2x3 – 24 – 25

5. a? +62. 6. 2(ay-bx)? + 2(ax – by). (1 + x)(1+x2)(1+33) 7. }{cx+y)s +(y+z)3 + (x+2)5 - 2025 +y5+:5)}. See Ex. 12, p. 30, Sec. IV.

x(x + 2y) 8.

9.

ab(1 – 2?) 10. (1–a)(1+02) + 2ab(a - b). x2 + 4xy – 2y?' 2(62 - a?)

(1 - a?)(1 - 62)(1-x2)

2

XXIII. 1 and 2 offer no difficulties.

3. a?c=x: – xyz, b2y=y: – xyz, c2z=z3 – 3xy ;
... aRx+b2y+c?z=23 + y3 +28 – 3.xyz=(x + y +z)(x2 + y2 +22 — ZY— X-- yz)
a2x + b2 y +cz

= x2 + y2 + 2 - xy – 2z - yz=a? + b2 +ca.
x+y+
4. Shew that xyz and x+y+z are equal with contrary signs.
5. Add a to each side of y+z+u=ax, then x+y+z+u=(a +1)x,

1 and

Similarly for each of the other three expressions.
a+1 X+ y +:+u'
6. Since a=

X - ?!

.. ax + ay=x -- y, and (1+aly=(1- a)x

3+y _ita

y_1+0

1+0 ; similarly

and y 1-a

1-C Whence 1+al+b 1+c

=1. 1-a 1-61-c ya 7. Multiply by xyz, and arrnge the terms so that x, y, z may be included in brackets in the same manner as a, b, c in the given expression.

8. The sum of the numerators of the four fractions when reduced to a common denominator is 433–3(a+b+c)82+2(ab+ac+bc)s- abc, which when reduced and the substitution for s made, the result is abc. 9. The given expression can be put under the form 62 +c2 - a?

-1+
c2 +0?–62 a? +62 - c?
+1+

-1=0,
2bc
2ac

2ab
(6—c)2—a” (c+a)2 – b? _(a−b)-c?=0.

-+
2bc
2ac

2ab
Whence (b-cta).(b-c-a)+(c+a+b)(c+a-b)+cla-b+c)(a-b-c)

= (a +c-6). (c(a--c)+b(c+a+b) - a(6-c+a)},

= (a +c-b). {62 – (a —c)2}=(a +c-6)(6+a – c)(b + c-a)=0. 10. First add the three fractions, then as the sum is equal to unity, the numerator of the sum is equal to the common denominator. Whence there results a3 + b3 + c3 – 3abc=0, and a3 + b3 + c3 = 3abc,

or (a + b + c)(a2 +62 +62 - ab - ac – bc)=0. Taking a +b+c=0, then a2 +62 - c? = - 2ab, a? + c2 – 1,2 = - 2ac,

and 6% +0° - a2 = -2bc. Whence

72

c2

ila?

72 c?)
+
62 +62 - a2 .a? +02 - 12.a? +62 - 02 2 l uc al

15a3 +63 +03

abc
1 3abc 3

2 abc 2
11. Here (a? – bc)(12 – ac)(c? – ab) = 0,
or a2b2c2 – 6%c3 - a8c3 +abc* - a3b3 +6+ac+a+bc – a2/2-2_O,
.', a368 +a3c3 +68c3 =abc4 +acb+ + bca“,

1 1 1 a3 + b3 + c3 and dividing these equals by a3b3c", there results

+ +

63 cs alca

or

+

+

+ ac