3. Subtract

(a+y)x3—(2a2—y2)x2-2xy3 from a3y — (2x2 — a2)y2+ax3,

-

and arrange your answer in descending powers of y, collecting coefficients.

4. Arrange according to powers of x, collecting coefficients,

5. Simplify

3(a2 — 2x)(a− x3)+2x(a−bx)2.

(1) (3a−7b+4c) + (a−4b+2c) — (2a−3b+2c). (2) I—{1-(1—4x)}−{2−(−4+5x)}.

(3) a2 —(b2 —c2) — { b2 — (c2 — a2) } + { c2 — (b2 — a2)}•

(4) 3(x+a)(y—b)—{—a[b—(c–d)]}.

(5) 3a-1[b+{2a—(b−x)}].

(6) 3bx (c–d) —1{4ad—6(a—b)(c–d)}x.

(7) 4{(2ab2c1)1 — 2ab2c(2ab2c5)3}.

(8) (x+y)2—2x(3x+2y) — (y−x)(−x+y).

(9)

3a—{2a—(3a—b)2} +3a { 26—3a–62}.

a{26-3a-1}.

2

(10) ¿—m−3 [ { /− m + lm (} + =) } * −4] ·
l-m-3 {1-m

2x3

(11) 1− (1 − (1 + x)3) + 3 { 2 − (x − 1)°—2x2 } .

(12) 2{a(a—b) — (a+b)2 } (a2 — 2b2).
(13) 3b{a(a-b)(3a+b) — (a−b)3}.

6. Prove the following equivalents:
(1) x(a2 — ax— x2)+ax(x−2b—3c)

-alax-2bx-3cx)=x3.

3

(2) a2-{2(a+b)+c}c+2a{ 1 (a+b+c)—1b}

=2a(a−1c)+1b(a+2b)−(b+c)2.

(3) (p+q+r)2− (p2 + q2 + r2)=p(q+r)+q(p+r) +r(p+q).

m

11. What are the meanings of am, am-", an, and amn?

VI. GREATEST COMMON MEASURE.

1. Find the G.C.M. of

(1) 12a2x31 and 14a3xy3.

(2) a2x2 - 2ax, a2x3 — 3a2.

(3) 2ax-2ay, 3bx—3by.

(4) 8a31⁄23+12a3r2 and 12+18ax1.
(5) 9x2-25 and 9x2+3x-20.
(6) 12x-1082 and 16x1-48x3.
(7) 4x3 —xy2 and 4x3 — 10x2y+4xy2.

(8) aa— 2a3x+a2x2 and aa — 2a2x2 + xa.
(9) a-x, and a3+ a2x-ax2- x3

(10) 6a6-6a3x3 and 8a-8a3x.

(11) 6a2+7ax-3x2 and 6a2+11ax+3x2.

(12) 3x2+16x-35 and 5x2+33x— 14.

(13) бa3 −6a2y+2ay2 — 233 and 12a2 — 15ay+3y3.

(14) 7x2-12x+5 and 2x3+x2−8x+5.

(15) 2a3 +2a2y—3ay2 —373 and 3a2+ay—2y2. (16) · 12x6 +213 — 2 and 16x7 — 12x1 — 10x.

(17) x3+3x2+3x+2 and x3—x2 —x—2.

(18) 2x2+IIx3 — 13x2 — 99x—45 and 2x3-7x2
-46x-21.

(19) 3x3-22x-15 and 5x+x3-54x2 + 18x.
(20) 5x3-12x2+19x-6 and 3x-2323+43x2

-51x.

(21) 3x2+7x3-15x2+9x-2 and 6x3+23x2-39x

+14.

(22) x1-8x3+21x2- 20x+4 and 213 — 12x2+21x

- IO.

(23) j1+m2y2+m1 and y1+my3 — m3y — m2. (24) 325-1023+15x-8 and 2x1 −4x2 +2. (25) x3-3x2+7x−21 and 2x1+19x2+35.

(26) x1+5x3-7x2-9x-10 and 2x1—4x3-3x2

+8x-4

(27) xa +x3−2x2+3x−3 and x3— 413 — 2x2+ 3x

+2.

(28) 3-8x+3 and x6+3x+x+3.

(29) a2-4x2+12x-9 and a2+2a−4x2+8x−3.

2. Explain what is meant by greatest common

measure.

3. Prove the rule for finding the G.C.M. of two quantities, when neither of them has any simple factor.

VII. LEAST COMMON MULTIPLE.

1. Find the L.C.M. of

(1) axy and a(xy—y2).

(2) x2(x−y) and x(x—y)2.
(3) 2a2+2axy and axy+x22.

(4) a2-ab, a2b, ab2, ab-b2.

(5) b2+ bxy and bxy — x2y2.

(6) 2a2(a+x), 4ax(a−x), 6x2(a+x).

(7) xy, x-y, y3 — x2y.

(8) 6ax3+6bx3, a2x2—b2x2, 15x.

(9) 3x2+6xy2 and 2x2— 8y2.

(10) 4(a2-a) and 6(a2+a).

(11) x, x2, (x+ 1)2, x2 — 1, (x− 1)2.

(12) x3+y3, x3− y3, 3(x−1).

(13) 3(1 − x2), 6(1−x)2, 5(x+x2).

(14) a−x, a+x, a2 —x2, a3 +x3.

(15) 4(1−x)2, 8(1−x), 8(1+x), 4(1+x2).
(16) 4ab(a2-b2)3 and 6a3b2-6a2b3.
(17) a−I,x+x−2, x2−52–6.

(18) a2 — 1, a2+2a−3, a3—7a2+6a.

(19) x3+x2y+xy2+y3 and x3 — x2y +xy2 — y3.
(20) 2x3+5x2y+5xy2+3y3 and 2x3+x2y—xy2
+313.

(21) x3-8x2+7, x3+7x-8.

(22) 3x3—15ax2+a2x−5a3 and 6xa—25a2x2 — 9aa.

2. Find the G.C.M. and the L.C.M. of

(1) (2a2bx+2abx2)2 and (6a2b2—6b2x2)3.

(2) 2x3 —9x2y+11xy2 — 3μ3 and 4x3—4x2y—5xy2 +373.

3. Find all the quantities whose L.C.M. is 3 — 4xy2.

4. Explain what is meant by a common multiple, and what by the least common multiple of two or more quantities.

5. State how the least common multiples of quantities may most conveniently be found when the quantities are resolvable into factors.

6. Establish a rule for finding the L.C.M. of two quantities when their factors are not discoverable.

7. Shew that every common multiple of two quantities is also a multiple of their L.C.M.