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9. (a+b+c)3+a3+b3+c3=(a+b)3+(b+c)3+(c+a)3+6abc. 10. (a—b)3+(b− c)3 + (c—a)3 = 3(a—b)(b—c)(c—a).

11. {(a—b)2+(b—c)3+(c—a)3}2 = 2 {(a—b)*+(b—c)*+(c—a)*}. 12. {(a—b)3+(b—c)2+(c—a)2}3—54(a—b)2(b—c)2(c—a)3 =2(a+b—2c)2(b+c−2a)2 (c+a—26)3.

13. (a—b)*+(b—c)*+(c—a)* = 2{(a—b)2(c—a)2+(a—b)2(b—c)2 +(b−c)2(c—a)3} = 2(a2+b2+c2—ab—bc—ac)3.

14. (b3+c2—a2+ab+be+ca)2 (c2—b3)+(c2+a2—b2+ab+be+ca)2(aa—c3) +(a2+b2—c2+ab+bc+ca) = (a2—b3)(b2—c2)(c2—a3).

15. a2(b+c)3+b2(c+a)3+c2(a+b)2+abc(a+b+c) +(a2+b2+c3)(bc+ca+ab)=(a+b+c)(b+c)(c+a)(a+b).

16. {2bc(a-b)-(b3+c2—a3) (a—c)}2+(a—c)3 (b+c-a)(c+a—b)

(a+b−c)(a+b+c)=4abc{abc—(b+c−a)(c+a—b)(a+b−c)}.
17. a'(b+c)+b (a+c)*+c‘(a+b)* = 2b3c2 (a+b)2 (a+c)2
+2c2a2(b+c)2(b+a)2+2a2b2(a+c)2(c+b)2—16a2b3c2(bc+ac+ab). ·
18. (a+b)2(b+c)2(c+a)2+2a2b3c2—a1(b+c)3—b1(c+a)3—c‘(a+b)*
=2(ab+be+ca)3.

XIII.

Shew the identity in value of the following sets of expressions:1. (a—b)3+(c—a)(c—b), (b−c)3+(a—b)(a—c) and

(c—a)2+(b—a)(b—c).

2. (a+b)2+(b+c)3+(c+a)3—(a3+b2+c2), 3(a2+b2+c3)-{(a—b)3+(b−c)2+(c—a)3}, and {a-b)+(b—c)2+(c—a)2+6(ab+ac+bc)}.

3. ac(a—c)+ab(b—a)+bc(c—b), a3(b−c)+b2(c—a)+c2(a—b), and (a—b)(b—c)(c—a).

4. (a+b+c)(ab+ac+bc)—abc, and (a+b)(b+c)(c+a).

5. a(b-c)+b(c—a)3+c(a—b)3, and (a-b)(b—c)(c—a)(a+b+c). 6. a(b+c)2+b(c+a)+c(a+b)3—(a+b)(b+c)(c+a), and a(b+c-a)3+b(c+a—b)2+c(a+b—c)2+(a+b—c)(b+c—a)(c+a—b). 7. }{(a+b—c)(a+b)+(c+a−b)(a+c)+(b+c−a)(b+c)}. }{(a+b+c)2+(a+b−c)2+(a+c—b)3+(b+c—a)2}, and 2(ab+ac+bc)—(a+b—c)(a+c—b)—(b+c—a)(a+b—c)

—(a+c—b)(b+c—a).

8. {a(a—b)+b(b−c)+c(c—a)}(a+b+c), and
{(a+b)+(b+c)3+(c+a)3—3(a+b)(b+c)(c+a).

9. (a3+b3+c2)3+2(bc+ac+ab)3—3(a2+b2+c2)(bc+ac+ab)3,
(a3—bc)3+(b2—ac)3+(c2—ab)3—3(a3—bc)(b2—ac)(c2—ab), and
(a2+2bc)3+(b2+2ac)3+(c2+2ab)3—3(a3+2bc)(b2+2ac)(c2+2ab).

10. (a+b+c)(a+b−c)(a+c—b)(b+c—a), 4a2b3—{c2—(a2+b2)}2, a2(b2+c3—a2)+b2(a3+c3—b3)+c3(a3 +b3—c3), 4(a2b2+a2c2+b2c2)—(a2+b2+c2)3, (a2+b3)3 — (a3—b3)3 — (a2+b2—c2)3.

and 2ab+2a3c3+2b3c2 — aˆ— õ1—c3.

XIV.

Verify the following equivalent expressions:-
:-

1. (1+x)2(1+y3)—(1+x2)(1+y)3 = 2(x—y)(1—xy).

2. (2x+3y)+(3x+2y)3=5(x+y) (7x2+11xy+7y2).

3. (a−b)(x − a)(x − b)+(b−c)(x− b)(x−c)+(c− a)(x — c)(x − a) — (a—b)(b — c)(c—a).

=

4. (x+b)(x+c)−(a+b+c)(x+b)+a3+ab+b2+3ax

=(x+a)-(a−b)c.

5. (xy — a2)2+(ay — bx)(ax — cy) = (ca — x2) (ab — y3) + (bc — a2)(xy — a2). 6. (a+b+c)(x+y+z)+(a+b−c)(x+y−z)+(b+c− a)(y+s−x) +(c+a−b)(z+x− y)=4(ax+by+cz).

7. (x+y+z―xyz)2+(xy+yz+zx−1)2 = (1+x)2(1+y2)(1+z2). 8. x(y+z)+y(x+z)2+z(x+y)2-4xyz=(y+z)(z+x)(x+y). 9. (1+xz)2(1+yz)2 — {(1−xz)(1—yz)+2xyz}2 =4(x+y—xy)(xyz3+xyz2+z).

10. (xyz+x2y—y2z+z2x)3+(xyz+xy2+yz3—zx2)3

=(x2+y3)(y2+z3)(x2+x2).

11. (x2-1)(y2-1)(x2-1)+(x+yz)(y+xz)(z+xy)

=(xyz+1)(x2+y2+z2+2xyz−1).

12. (y2—zx)(z2—xy)+(z2—xy)(x2—yz)+(x2—yz)(y2—zx) —(x2—yz)3—(y2—zx)3 — (z2—xy)2 = (x+y+z)(3xyz—x3—y3—¿3). 13. (x2+y3)(x2+x2)(y—z)+(x2+y3)(x2+y3)(x-x) +(x2+z3)(y2+z2)(x—y) = x (y-z)+y'(z—x)+z1(x—y). 14. (y-z)+(-x)+(xy) =5(y—z)(x—x)(x—y){x2+y2+z2—yz—zx—xy}.

15. (y—z)+(x− x)°+(x− y)° = 3(y — z)3 (z — x)2(x − y)3 +2(x2+y3+s3 − xy — yz — xz)3.

XV.

1. If A= ax+by, and B-bx-ay, shew that a▲3+bAB+cB3 = (a+b+c)b2x2+(a2—2ac+b2)bxy+a2cy3.

2. If A-a-bc, B-b-ac, C=c-ab; prove that (A-BC)bc (B-AC)ac = (C2-AB)ab=abc(a+b+c)(A+B+C). 3. If a3+b2+c3 = A, ab+ac+bc = B; shew that

A3+2B3—3AB2 = (a3+b3+c3—3abc).

4. Given a+b+c+d=A, a+b—c-d=B, a−b+c-d= C, a-b-c+d=D;

shew that AB(A2+B2) = CD(C2+D3), if ab(a2+b2)=cd(c2+d2).

5. If a+b+c+d=0,A= bcd, B = cda, C= dab, D=abc; then shall BCD+CDA+DAB+ABC=0.

6. If bc-d=A, ca-e2 B, ab-f2- C, ef-ad-D, fd-be = E, de-cf-F; then shall

AD2+BE2+CF2 = ABC-2DEF—(Aa+eE+ƒF).

7. If X-ax+by+cz, Y=cx+by+az, and Z=bx+ay+cz, prove that (a3+b+c3—3abc)(x3+y3+z3—3xyz) = X3+Y3+Z3—3XYZ.

8. If X-ax+cy+bz, Y=cx+by+az, Z=bx+oy+cz; shew that X+Y+Z3- YZ-ZX-XY

=(a+b2+c2-be-ac-ab)(x2+y2+s2—y%-zc-xy).

9. If the six equivalents, a=xX, b=yY, c=zZ, 2A=yZ+zY, 2B=2X+xZ, 2C=xY+yX, be simultaneously true, then shall aA2+bB+cC2 = 2ABC+abc.

XVI.

If 28=a+b+c, 203 = a2+b2+c3, and 203 = a3+b3+; shew the truth of the following equivalents :

1. s2+(8—a)2+(8—b)2 +(8—c)2 = 203. 2. (s—a)3+(s—b)3+(8—c)3+3abc = s3. 3. (s—a)(s—b)(s—c) = 83—80-abc. 4. (b+c)s(8-a)+a(s—b)(8—c)-2bcs =(c+a)s(s—b)+b(s—c)(s—a)—2cas =(a+b)8(8−c)+c(s—a)(s—b)—2abs.

5. 8(8—b)(8—c)+s(s—c)(s—a)+8(8—a) (8—3) —(s—b)(s—c)(s—a) = abc.

6. {(s—a)(s—b)(s—c)}2+{8(8—b)(8—c)}2+{s(8—c)(8—a)}}'

+{s(s—a)(s—b)}2+2802s(s—a)(s—b)(s—c) = a2b2c2.

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8. (8—a)(8—b)+(8—b)(8−c)+(8—c)(s—a) = s3 —o3.

9. (o3—a3)(8—a)+(03—b3) (8—b)+(03—c3) (s—c) = a*+b‘+c‘—803.

XVII.

1. If bz-cy=p, cx-az-q, ay-bx=r; then ap+bq+cr=0.

2. If a=y+x-2x, b=z+x-2y, c=x+y−2%; find the value of +c+2bc-a3 in terms of x, y, z.

3. If 3x=-p+2q+2r, 3y=2p−q+2r, 3z=2p+2q-r; show that x2+y2+z2=p2+q3+r3, and xy+yz+xz=pq+pr+qr.

4. If p=b+c+d—a, q=a+c+d—b,r=a+b+d—c, 8=a+b+c-d; shew that p(q+pr+ps)+q(a+qs+rs)=4(ab+ac+ad+be+bd+cd). 5. If (x—y)z2 = c3, (y—z)x2 = a3, (x—z)y2 = b3, (x—y)(y—z)(z—x) = 3abc, then shall a3+63+c3—3abc=0.

6. If 2a=x+%, 2b=x+x, 2c=x+y, find the value of the expression a+b+c-261c23-2c3a3-2ab in terms of x, y, z, and express (x+y+z)(xy+xz+yz)-xyz in the form of factors involving a, b, c.

7. y3+z+m(y + z) = =3+x3+m(z+x) = x2+y3+m(x+y), x, y, z being unequal, then each expression is equal to 2xyz.

8. If a2=y+%, b2=x+z, c2=y+z, and 28=a+b+c, shew that s(8—a)(8—b)(8—0) = 4(xy+xz+yz).

9. Shew that (a+b+c−d)(a+b+d—c)(a+c+d—b)(b+c+d—a) =16(8-a)(8—b)(8—c)(s—d); if 28 = a+b+c+d.

10. If a+b+c=38, shew that (8-a)+(s—b)+(8-0)" =2{(8—b)3 (8—c)3+(s—c)3(8—a)3 +(8—a)2(s—b)3}.

11. If x+y=p and xy=q, find x2+y3, x2+y3, x+y, &c., in terms

of p and q.

12. If a2+b2=c2, the product (a+b+c)(a+b−c)(a+c—b)(b+c—a) is equivalent to 4a2b3.

13. If xyz=a, y2-xz=b, and x-xy=c; then shall

x+y+z3-3xyz= ax+by+cz.

XVIII.

1. If x(1+y)= 1 and y(1+3)=2, then -8=1+x+2x2+4x2+.... 2. If ay+bxa, by-axb, then x2+y=1.

3. If xy=x(x+y-z), then shall (x-x)=yz.

4. If x+y+z—xyz=2, then shall (1—x)2 = (1—xy)(1—xz).

5. If b(bx2+a3y) = a(ay2+b2%), then shall bx+ay = ab, and ay = bx. 6. If (a+b-c-d)x-cd-ab, then (a+x)(b+x)=(c+x)(d+x). 7. If ax+by=1, then ab(x+y)+(a3+b2)xy+(a—b)(x—y) = 1. 8. If (a-bc)x+(b3—ca)y+(c2—ab)z=0 and x+y+z = 0, prove that ax+by+cz=0.

9. If (a+bc)2(1—a2) = (b+ac)3 ( 1 —b3), then a2+b2+c2+2abc = 1. 10. If x3+y3=z2, prove that (x3+z3)3y3+(x3—y3)3z3 = (y3+z3)31⁄23. 11. If a+b+c=0, then 6(a+b+c3)=5(a3+b3+c3)(a2+b2+c2), 4(a2+b2+c2)=7(a+b+c1)abc, and (a3+b3+c3)3 = 27a3b3c3.

=

12. If ax+by-c3=0, ay2+b2x-c-0 and x+y-c=0, then bac. 13. If (2a—3y)y = (≈—x)2 and (2a—3z)z = (x—y)3, then shall x+y+za and (2a-3x)=(y-z).

14. Ifa+b+c=0, and a (by+cz—ax)=b(cz+ax—by) = c(ax+by—cz), then will x+y+2=0.

15. If (by—cx)2 = (b2—ac)(y3—cz), then shall

(bx-ay)=(b3-ac)(x3-az).

16. If a+b+c+d=0, then shall a+b+c+d'+4abcd

= 2{(ab—cd)2+(ac—bd)2+(ad+bc)2}.

=

17. If (yz—x)3+(zx—y)3+xy—z)3 = (yz—x)(zx—y)(xy—z)+4, then x2+y2+z3 = xyz+4; and conversely.

XIX.

1. If a be greater than b, a”—b” is greater than nb"—1(a—b) but less than na®-1(a—b).

2. Write the pth term of the quotient of a-bTM when divided by a"-b".

3. Show that a—a" is divisible by a+1, when m―n is even ; a”+a" is divisible by a+1 when m―n is odd; and that a"-a" is divisible by a-1 when m―n is even or odd.

4. Prove that a-1 is always divisible by a"-1 and by 2"-1; and write the middle terms of the quotients when m is odd, and when n is even.

5. Determine the number of factors by which "P-1 is divisible,

when m, n, p are each odd and even numbers; and state the number of terms in each quotient.

6. Shew that 26=1+(1+2)(1+22)(1+2′)(1+2°)(1+2:6)(1+2).

7. Show that x”—na”—1x+(n−1)a" is divisible by (x-a)', without performing the operation; and write the exact quotient of x-5ax+4a2 divided by (-a)3.

8. Prove that (a"—1—b"—1)(a"—b")(a"+1—b"+1) is divisible by (a—b)(a2—b3)(a3—b3).

XX.

1. Shew by means of division if x2+acy+y+bx+cy is resolvable into two rational factors, that abc = b2+c2.

2. Write the coefficient of x' in the product of x—ax3+a2x2—a3x+a*, x2+ax+a and x-ax+a3.

3. How many factors of the form x+a, and of the form x-b, are contained in the product -11x+33x3-11x3-154x+120?

4. Divide +px2+px+1 and x2+px2+qx3+qx2+px+1 by x+1 respectively, employing any artifice to save the trouble of formal division.

5. What is the criterion by which it is known that an Algebraical polynomial with integral coefficients, and arranged according to descending powers of x, is divisible (1) by x+1 and (2) by x-1? Shew that 2+3x1—x3—x2-2 is divisible both by x+1 and by x-1, and write the quotient.

6. Shew that —px3+qx3-rx+8 is exactly divisible by x-a, if a1—pa3+qa3—ra+8=0.

7. Divide x-px1+qx3—rx2+sx-t by x-a as far as five terms of the quotient, and prove the correctness of the quotient by multiplying the divisor and quotient together and adding the remainder.

Also write the quotient and remainder, (if any)

when 2-8x+12x3-18x2+20x-30 is divided by x-4.

8. Shew when a1+a1x+α„x2+azx3+

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+a„x”, can be divided

+P, be divisible by x-a, with

out remainder, shew that a"+p1a"-'+P2a”-2+ . . . . +P„ is equal to

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Ex. x+5x+6x*+5x2+1=(x2+2)*—3(x2+2)+9(x2+2)—9.

11. If there be a series of factors x+y, x+y−1, x+y−2, &c., and if the second factor be written in the forms (x-1)+y and x+(y-1), then the product of the first two factors will take the form of x(x-1)+2xy+y(y−1).

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