10. The third power of - 1 - 3 is +1; hence this expression exbibits the two / -3 symbolic cube roots of +1. And the third power of +1EV - 3 is – 1, and the expression gives the two symbolic cube roots of -1. And -1+V-3_-]+V-3-1-V-3 -1-1-3-1-7-3 2 2 2 XXVIII. The first eight examples offer no difficulty beyond the exact performance of the operations indicated. 9. This may be shown in two ways: by extracting the square roots of each quantity and taking the sum of the results, and by squaring each side of the expression, and showing that the result when reduced is 32 +68=100, an obvious identity. X Х Х XXIX. y - XV-1 y - 28 -1 2x – YV-1 (- YV-1)(y+31-1 ✓ (x? +y?)V-1 (2c – yd - 1) - (y2 + x)2-(++y*)). . 3 XXXIII. 1. 3, 3/3, 9, 9V 3. 2. 75, 375–3, 5025, 28125/3. 3. 12, 24V3, 144, 288/3. 4. , V 5, 6, 34V 5. 5. , 1977, 1977. 6. 72, 144/2, 576, 1152x2. 7. 9V9, 81, 243V3, 7298/9. 8. IV36, f, V6, si V36. 9. 25V25, 625, 3125V5, 15625/25. 10. 7+4V3, 26–1573, 97+563, 362+ 209/3. 11. 5+2/6, 11/2+9/3, 49+2016, 109/2£89V 3. 12. 77 +30/6, 6552+531/3, 11329+ 4620/6, 98225/2+80187 V3. XXXIV. (5V 5)!=(5.5)}=51.57=(525)!=(125)+ or V125, (5/5)+=(5.51}}=51.57=(52.5)&=(125), or V195. The square roots of the last four examples can be found as the example in tho note, p. 15. The roots are respectively 3+V3, V3+1, 3/3 - 2/2, 2+31 -1. XXXV. - 3V75=-3V (25 x 3)=-3X5V 3=-1513 .. 3V 147 - 37 75-3Vf=(21-15-1)/3=5V 3. then x=a+b, but a3 +68=2V5, and 3ab=3, .. =2V5+3x, and x3 – 3x=215. XXXVI. Since (Va+V)(Va-Vb)=a-b, and (avx+y)(ava-Vy)=a’x-b?y; it is obvious that Va + V6 may be made rational by multiplying it by Va-Vb; and va-vb by multiplying it by Va+Vb. Similarly for the factors ava+bVy and avx-ONy. When there are three terms, Va+16+Wc, since {(Va+Vb)+Nc}•{(Va+b) – Vc}=(Va+vb)*_c=(n+6–c) +2 Vab, and {(a +6–c)+2aV6}. {(a+b-c) – 2Vab}=(a +6–c)2 - 4ab. And two factors will be required in general to render such a denominator a rational quantity. 8-5V2 8–512 3+212 24 +12 - 20 1. -4+ V2. 3-212-3-212X3+272 9-8 83 8 16. Hare (5.12)&= {} 102 107 (-03375) := {31059 33.53 ) } 15 3 10.103 2.108 2x10 20 103 101 101 1 3 8+ 101 19 1 1 becomes 2 192 (101)3= { } { 20-1} 102 1. -+ + XXXVII. 1. These reductions offer no difficulty, as V245+ V75 V245 – 175_7V5+5V3, 715-513 50+12V15 50 - 12V15 2 = 50. + XXXVIII. 1. The expressions 2, 3, 4, 5, 6ă, are respectively equivalent to (230)37, (32033', (41570%, (512), (619)**; and by expanding the powers 230, 320, 415, 513, 610, their numerical values will be known and the given expressions can be arranged in the order of magnitude. 2 2 2. V3_203 3 Х V6. (3) )' ( 32 4 3 )} 3 12 V2 Х 1 1 1 999 The extractions and divisions in each example are to be carried out to four places of decimals. 3. 4 is greater or less than V5. According as 4 51 53, raising each to the 12th power. 125, Similarly 2l is less than 31. 312 is greater than 2 V3. (52)# is less than (5')}, and 525–51.25)... 17 is greater than 23/3. is less than V3 V5 22+2V57, by squaring each side. 2770 5+2V57, subtracting 17. 280 25+ 20757 +228, but 280 is obv ly less than 25+20V57 + 228, .. V10+ 17 is less than / 19+V3. (2) V2+17 is greater than V 3+1 5. (3) V5+V 14 is greater than V 3+3V2. (4) V 6-V 5 is greater than V 8-17. (6) 23 +52 is less than 3. 5. If every two of the three lines be greater than the third, a triangle can be formed with the three lines. 6. It is obvious that if the sides of the equilateral triangle, the square, the pentagon, be given in magnitude, the areas of these regular figures can be determined, and consequently the surfaces of each of the five regular solids. The surface of the tetrahedron is equal to the area of four equal and equilateral triangles. The surface of the hexahedron is the area of six equal squares. The surface of the octahedron is equal to the areas of eight equal and equilateral triangles. The surface of the dodecahedron is equal to the area of twelve equal equilateral and equiangular pentagons. The surface of the icosahedron is equal to the area of twenty equal and equilateral triangles. A sphere can be inscribed within, and another circumscribed about, each of the five regular solids, in the same manner as a circle can be inscribed within and another circumscribed about each of the plane regular figures, the equilateral triangle, the square, the pentagon. Suppose a sphere described about a regular tetrahedron, the surface of the sphere will pass through the four points of the trihedral angles of the tetrahedron. If lines be supposed to be drawn from these points to the centre of the circumscribing sphere, these lines will be each equal to the radius of the sphere. Hence the tetrahedron may be conceived to be divided into four equal triangular pyramids, whose common vertices are at the centre of the sphere, the three edges of each, radii of the sphere, and their four bases the equilateral triangles which make up the surface of the tetrahedron. Since the content of any pyramid is equal to one-third of the content of a sm on the same base and of the same altitude, the contents of the four equal pyramids can be found, and the content, or volume, of the tetrahedron determined. In a similar manner it may be shown that: The volume of the hexahedron is equal to the volumes of six equal pyramids whose bases are equal squares. The volume of the octahedron is equal to the volumes of eight equal pyramids whose bases are equal equilateral triangles. The volume of the dodecahedron is equal to the volumes of twelve equal pyramids whose bases are equal equilateral and equiangular pentagops. The volume of the icosahedron is equal to the volumes of twenty equal pyramids whose bases are equal equilateral triangles. a 3. Since c= (1 - 62)# +6(1 - a2)}; ..c-a(1 – 62)+=611 - a2)}, cata? – b2=2ac(1 –62), .. 4a27*c= 2a?c? +26c+ 2a2b2-a4-64 - Ch. x(a? - y2)!=a? - y(a? — 242)!, by squaring each side, and 2y(a? – xa)=a2.- (? - Yo), squaring again; whence a4 – 2a? (x2 + y2)+(x2 + y2)=0 .. a? – (? +y?)=0 and 22 + y2 =a?. and 3+ y + a + 2x+y} – 2atxi - 2ały=0 {(x+y=a)? + a[x+y)+ ala – x–y)+al+y-a} =(x+y-a)? + xa(x+y) = 2(x2 + y2 +a%). The truth of this also may be verified by numerical examples, az, for instance, when (16)++(25)' = (81)+. 8. If (a + b +c+dł)!=xt + y + zi, Then a +61 +c+d+=x+y+z+ 2(xy)t +2(xz)} + 2(yz)", and x+y+z=a, 2(xy).=, 2(xz)t = ct, 2(yz)t=di. .. 2a(bcd)t=16xyz(x+y+z). But 4xy =b, 4xz=c, 4yz=d, .. bc + bd + cd=16(xoyz+yəxz+zoxy)=16xyz(x+y+z). Hence, therefore, 2a(bcd)t=bc+bd+cd. 9. Let {r+(72 +9°)}}}=a,{r-(+9?)"}}=b; then x:=(a+b):=a3 + b3 +3ab(a+)=2r- 3qx ; .. 28 +3qx - 2r=0. 10. Find y and z in terms of x, and substitute these values in 2-+y'+x=1, whence x is known, and y, z by substitution. Lastly, substitute these values of *, y, x in ax? + by2 +cz3. XL. 4. Since alb - c)- c(b+c)2=C; a (6+c)? at bto and 16-c)?' 9 |