The student may try to resolve a358 +a%c3 + b3c3 — abc(a + b3 + c3)

and a%b3 + b3c8 +a%c3 + abc(as + b3 + c3 + 2abc), each into three factors.

or 3.

ac

1

, Y-'=

16(a+b)222 _

XXIV.

3(x - a)(x-7)(x —c) 1. The difference is

(.1-a)(x - 5)(x - C)

1 2. Here xyz=1 and y==

and x2=

y
1
1
1
1

1+x+xz
then
+

=1. 1+x+;*1+3+1+1+y+"1+x+uz+1+x+r+1+x+3x1+x+rz 3. The substitutions for x, y, z being made, the expression becomes ab bc

2abc (a+c)(6+c)(a +c)(a +0) 16+c)(a +b) (a+b)(6+c)(c+a)' which when reduced gives the required result.

4. Only requires the substitutions to be made for x and y.

5. The truth of this may be directly shewn by simply substituting for x and y their values in terms of a and b.

Or indirectly, by assuming the result as true, namely, 13+y=y3 +®, then 33 – y=x – y and a2 + xy + y2=1, and if for x and y their values be substituted, the sum be equal to unity, and the truth is proved. 6. 3-a= a? + 2ab + 512

a: + 2ab - 362 4(a+b)

4(a - b) then (x – a)? +(y - 3)2 =

16(a + b)2 7. The given expression is reducible to (x+y+z)(xy+xz+y=)=0. 8. First,

су

whence ad.c(d+y)=bcy(b +x)

0(b + x) d(d+y)' and bacy – adx== xy(ad – bc), then dividing these equals by bdcy bc ad xylad – bc) dx by

bury 6 a 9. If ab+cd, ac+bd, ad + bc be denoted by p, q, r respectively, it may be easily shewn from the equality of the first and second fractions, that p+q+r=0. The same condition may be found from the equality of the second and third fractions, whence P+9_P+r_9+r.

-1.

9 9 10. Let ab +ac - bc=p, bc + ac – ab=, ab + bc - ac=r,

then Pl_Pre_9, whence p=q and q=r,

9 P
and restoring the values of p, q, r, ab + ac – bc=bc +ac - ab, .. a=C,

bc+ac - ab=ab+bc - ac, ..c=b, hence a=b=c.
11. The sum of these six fractions may be put under the form of
(z: - a)(x + a – b-c), (x – b)(x+b-a-c), (x – c)(x+c-a-b)

+ (a - b)(a - c) (6 - c)(6-a) (c- a)(c-b) and their sum is – 2ab +2(a + b)c - 2c2

a

a

с

с

or a

b十

bd+y

3. See Elementary Arithmetic, p. 13, Sect. IX., Exercises VIII. 9–12. 5. The former is greater or less than the latter, according as x2 is greater or less

than y?.

3

a

a

с

с

a

1

26 6. is less than

if (a + b)(b+c) be less than 26(a + c) or if ac +12 a+c

(a + b)(b + c) be less than ab + bc, or if (a - b,c be less than (a - b)b, that is if c be less than b, and a-6 be positive or a greater than b. 7. x may have any integral value between 1 and 7. 2

2 3 2 8. If x+=>3, then x? +2> 3x and 1+

-1, x2

22 2

1 and 2? > 3x – 2, .. 2? +.. > 3(x+ -1), but x+ > 2.

2? +

-> 3. x2 9. If a>b then a-6> 0 and a2 - 2ab +63 >0,.'. a? +62 > 2ab, and+">2. 10. Let the expression a(6+c) +62(a +c) + c2(a+b) or its equivalent

ab(a + b) + ac(a + c) + bc 6+c) be divided by abc, 6

6
then + + + is the quotient, and +-> 2, +-> 2, +-> 2,
bb

b
6
..-+-+ +C> 6, or ab(a + b) + ac(a + c) + bc1b+c)> bbc.

b
11. Since a>b, b>c, and a>c..a-6> 0, b-c>0, and a-c>0,

(a - b)(b – c)(a–c)>0,
.. (a - b)(6 - c)(a -c) >0, and

abc
0
whence

> 0, and “+
b b

b 12. Let b+c-a=X, C + a-b=y, a +b-c

1 1 1 by substitution

y

[ + y + But x2 + y2 > 2xy, y2 + x2 > 2yr, 22 +22 > 2xz; and x?z + y2z> 2ry-, y2x+z2x > 2.xyz, x2y + z2 Y > 2xyz, also 3xyz = 3xyz. Hence by addition x?y + xy2 + x*: +xz2 + y2 + yz2 + 3xyz > 9xyz,

9
or (x + y + 2)(xy + x2 + yz)> 9xyz, whence xy + x2 + yz

x+y+z'
11 9
or -+-+->

ญ x + y +7
13. Let b+c-a=x, c+a-bey, a+b-c=z; then a +b+c=x + y +z,

and 2:1 = y +2, 2b =x+z, 2c=x + y.

7 Hence

ily+, 2+2, x + y
b+c-a
b a +b-c 2

Y Suppose x, y, z in order of magnitude, and shew the aggregate to be greater than 3.

6-1

1 0-1 1 1. Here a= andre

from which the required expres. b 1-C

6 1-C sion may be found.

bc 2. Let a

=b- -P; then a? - bc=pa and b? – ac=pb,

b

.. (a? – 62) + c(a - b)=(a - b)p and p=a+b+c.
bc
bc ac

ab

ah
:6
-b= from which a+b=

p=

3. Equate the values of 6 from each of the given expeessions.

hd 4. Here ax + a'=bx+u, bx+1=cx+é; from the first x=

a-t'

a-6

(L

a' - b
from the second, <= ; equate these values of x.

6
C-d

d-c. 5. Since + =0, then

and (a - b)(1+cd)=(1+a))(d – c), 1+ ab '1+cd

1+ull+c' ..a+abc - d. – abc=b+bad - c-ced, or asl + bc) – ((1+bc)=b(1 + ad) - c(1 + ad),

b-c

a-d or (a - d)(1 + bc)=(6 - c)(1+ad), ..

and

1+ bc ital' 1+ bc1+ad 6. 1

1 and

b-c

+

с

2

с

2

a

C

c

с

ay:

1

1. alc

e+ 1-6-+cz _ (1 – e)(1 - -); 7. First 1-=11+ez 1+ez

1+ ez

1+e+:+e: (1+e)(1+ )
next 1+x=1+
1+02 1+02

1+ez

1-2 1-elThirdly, by dividing the former by the latter,

1+x

1+el+7

a+by 8. If the value of y be substituted for y in := and the expression

c+dy' reducedl, and divided by b+c, the following result will be obtained : ad +1,2

ad +62 (c? +ad): dz

c2 + ad

-b and
; in which if

C,
6+0
b+c

ito the provision will hold good. 9. Here a2(h—c)<b>(a ---), clearing from fractions by multiplying each side

a-d b-d by (a - d) {-d), then a’ {b2 – (c+d)b+cd}=bz{a? – (c+d)a+cd}, whence (c+d)(b-a)ab=cd(12 - a2),

(c+d)ab=cil(b+a), dividing by b - a,

(c+d)ab_cd(b + a)
and

dividing by alcu,
abcd abcd
1 1 1

十二三二十一,
d

6
The converse requires only the reversion of the steps.
10. 6%"=1-a2x2=(1+ax)(1 - ax), a’x2=1-62.co=(1+ bx)(1 – 12),

12
1- ax

73 b + abic
whence

and
1+ 413
22

1 + ax
a 1+ b

a - alac and

and 1- buc

1- buc

7,3
therefore

=(a + b)(a? +1%).
1- bx
1 + ax

22
11.
X _(n+1)6 y_(a+b)a

and

X _ Y __(a+)(b.x - ay)
Y
6

b

y
_(a+b)a y_(a + b)b

b, and 2: _Y_(2+b)(aur - by);
b
y

b

ху
y (a+b)2 {ab(n2 + y2) – (a? +62)xy}
b

?y?
Bat
x2 - xy + y? _a? - ab +62

. . ab(x2 + y2)-(a? +62)xy=abxy – abxy=0. ху

ab

x(y* – 22)=0-c, then

=a-C

or

{

b-c

+

C-a

a-b

C-a

+

+ C-a

=

12.
z(x2 - y2)
a - 0,

z(x2 - y2) + x(y2 – –-)
Xy

xy

y2 y*(in2 – --)

2 – 22_202 =a - C, whence

XY 13. Put each fraction equal to p, then 2p=a+b, and 2p=c+d. 14. Let each expression be denoted by P, then 22+x=y8xzp, y2+x=x8y=P, 32+y-=zexyp,

2 and 2(x2 + y2 +-2)=pxyz(x2 + y2 +-2).. p=

xy 15. X==- – 1 s(a - b)y+ (a - b)=) f(b-c)x (b-cz)

Y

$(c-a)x, (c-a)y?

la- -c I?
.. 2+y+z=-

Ş(a-6)} + (a –c)2+b-a)}=&c
6-c

a-b
17. Let -=P, then x=(6-c)p and ax=a(b-c)p,
b.

similarly by=b(c-a)p, cz=c(a - b)p.
Hence by addition ax + by + cz=p(ab - ac+bc - ab+ac – be)=0.
Otherwise,
ax _ (b-c)a by _(c-a) (>_ (a - b)c ax+by+c2_0

=0.
ญ
y
Y

Y
18. Since ab +ac+bc=0, .', a! – bc=a? + ab +ac.
1 1

1
1

1 Hence

ab + ac + bc
+
-+-+

-0. a? - bc 12 - ac C2 - ab a+b+c

a+b+c abc 19. Here abc=

- abc=

ayz – x(x+y)(x+2) (x+y)(y + 2)(2+2)

(x+y)(x + 2)(y + 2) 1 (x+y)(x+2)(y+:) 22

and

(x + y)(x + 3)(y+). asl - bc) x?(+y+z) all – bc) x+y+By a similar process

and

may be shewn equal to the same

(1-ac) c(1 - ab) quantity. 20. x+yy+2-3 x + y - 2 ax+by - cz

add 1 to each and reduce ax +by - by +62 - 6 Y+:-* by + cz – ax

1

y+:-X to improper fractions, then

and
Y+2-3 by+cz - ax by +cz – ax

1
x+2-y
similarly

1 and

&+y-2

ax +cz - by a ax+by - cz Otherwise let the reciprocals of each =

==P, then ax +by – cz=P(x + y - x)) then adding the first and second ; next the first and

by + cz – ax=p(y+--~) third ; and, lastly, the second and third, and the ax+cz – by=p(x+- y)) results are, p =b, pra, p=c, and a=b=c.

C-a

C-a

C-a

C-a

EDITED BY

ROBERT POTTS, M.A., TRINITY COLLEGE, CAMBRIDGE,

HON. LL.D., WILLIAM AND MARY COLLEGE, VA., U.S.

EUCLID'S ELEMENTS OF GEOMETRY. 1. Euclid's Elements of Geometry, the University Edition, with Notes, Questions, and Geometrical Exercises, selected from the Cambridge Senate House and College Examination Papers, with Hints for Solution of the Exercises. Demy 8vo., pp. 520, 10s.

2. The School Edition, with_Notes, Geometrical Exercises, &c. 12mo., pp. 418, 48. 6d. Books I.-IV., 38. Books I.-III., 28. 6d.

, . Books I., II., 1s. 6d. Book I., 18.

The University Edition of Euclid's Elements was first published in 1845, and the first School Edition in 1846. Both Editions have been enlarged and improved from time to time, and the total sales of copies of the work up to the present year amount to a number very considerably above half-a-million.

In the year 1853, the Council of Education at Calcutta were pleased to order the introduction of these Editions of Euclid's Elements into the Schools and Colleges under their control in Bengal.

In the year 1860, a Translation of the Geometrical Exercises was made into the German Language, hy Hans H. Von Aller, with a Preface by Dr. Wittstein, and published at Hanover.

At the International Exhibition of 1862, in London, a Medal was awarded to R. Potts, “For the Excellence of his Works on Geometry.” Jury Awards, Class xxix., p. 313.

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Critical Remarks on the Editions of Euclid. “In my opinion Mr. Potts has made a valuable addition to Geometrical literature by his Editions of Euclid's Elements."—11'. Whewell, D.D., Master of Trinity College, Cambridge. (1848.).

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“Mr. Potts' Editions of Euclid's Geometry are characterized by a due appreciation of the spirit and exactness of the Greek Geometry, and an acquaintance with its history, as wel! as by a knowledge of the modern extensions of the Science. The Elements are given in such a form as to preserve entirely the spirit of the ancient reasoning, and having been extensively used in Colleges and Public Schools, cannot fail to have the effect of keeping up the study of Geometry in its original purity.”—J, Challis, M.A., Plumian Professor of Astronomy and Experimental Philosophy in the University of Cambridge.. (1818.)

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greater Portion of this part the Course was printed, and had for some time been in use in the Academy, a new Edition of Euclid's Elements, by Mr. Robert Potts, M.A., of Trinity College, Cambridge, which is likely to supersedé most others, to

" When