Dividing both expressions by x-1, we obtain as L. C. M.

28. Divide the first expression by 2a, and the second by 3: then
a2-5a+6) a* - 5a3+8a2- 8(a2+2
a4-5a3+6a2

Dividing both expressions by a -2, we find as L. C. M, 6a (a-2) (a-3) (a3 - 3a2+2a+4).

29. x3-x2+x+3)x2 + x3-3x2- x+2(x+2

x3 + x2+3x

Dividing the two expressions by x+1, we obtain as L. C. M.

(x+1)(x2-2x+3) (x3-3x+2).

2x+2

2x+2

10a4+6a3-14a3- 6a+4) a5+ 4a3- 6a2

10a4+10a3-10a2-10a

4a3 4a2+ 4a+4 4a3 4a2+ 4a+4

Dividing both expressions by a3+a2 — a − 1, we obtain for L. C. M. (a3 + a2 — a − 1) (a3 — a2 — 4a+4) (a3 − a +2).

Hence L. C. M. of the first two expressions is

(x+1)(x+5) (x+6)=x3+12x2+41x+30.

Then to find L. C. M. of this result and the third expression, we have x2+11x+30) x3 +12x2+41x+30(x+1

x3+11x2+30x

x2+11x+30
x2+11x+30

this shews that the third expression also divides into the L. C. M. without remainder; hence the L. C. M. of the three is (x+1)(x+5) (x+6).

Hence L. C. M. of the first two expressions is (x-3) (x+5) (x − 2) or x3-19x+30; then taking this with the third expression, we have

Hence the complete L. C. M. is (x+5) (x − 3) (x − 2) (x2+1).

Hence L. C. M. of the first two is (x − 2) (x+2) (x − 5) or x3 – 5x2 - 4x+20.

Now take this with the third expression:

x3- 5x2-4x+20) x3- 5x2+4x-20 (1

x35x2-4x+20

8)8x-40

34.

Hence the complete L. c. M. is (x-2) (x+2) (x − 5) (x2+4).

6x2-7x-20) 9x2-16

2

18x2-32 (3

18x2-21x-60

7)21x+28

Hence the L. C. M. of the first two expressions is

(3x+4) (2x-5) (3x-4), or 18x3 - 45x2 - 32x+80,

Now take this with the third expression:

6x3- 13x2-9x+10) 18x3-45x2 - 32x+80 (3 18x3-39x2-27x+30

6x25x+50) 6x3-13x2- 9x+ 10(-x+3
6x35x250x

- 18x2+41x+ 10
-18x2-15x+150

28)56x-140

2x 5) 6x2+5x-50 (3x+10

6x2-15x

20x-50

Hence the complete L. c. M. is (2x-5) (3x+4) (3x-4) (3x2+x − 2).

20x-50

(2x − 1) (x2+3) (x2 − x − 1) = 2x3 – 3x1 + 5x3- 8x2 - 3x+3.

Now take this and the other expression;

x1 − x3 + 2x2 − 3x-3) 2x5 -3x2+5x3- 8x2 - 3x+3(2x − 1

2x5-2x+4x3- 6x2 - 6x

this shews that the third expression is contained in the L. C. M. already obtained.

Again multiply (1) by 5; then 5x+10y+15z=100;