Find LCM of the following algebraic expressions.
(i) 2x2-18 y2, 5 x2y+15 xy2, x3+27y3
(ii) (x+4)2 (x-3)3, (x-1) (x+4) (x-3)2
(iii) 10 (9x2+6xy+y2) , 12 (3x2-5xy-2y2), 14 (6x4+2x3)
(iv) 3(a-1), 2(a - 1)2 , (a2-1)
(i) Solution :
2x2 - 18 y2, 5 x2y+15 xy2, x3+27y3
2x2 - 18 y2 = 2(x2- 9y2)
= 2(x2-(3y)2)
2x2 - 18 y2 = 2(x+3y) (x-3y) ----(1)
5x2y+15x = 5xy(x+3y) ----(2)
x3+27y3 = x3+(3y)3
= (x+3y) (x2+x(3y)+(3y)2)
= (x+3y) (x2+3xy+9y2)
= 2(x+3y) 5 ⋅ x ⋅ y ⋅ (x2+3xy+9y2)
= 10xy(x + 3y) (x2+3xy+9y2)
So, the required least common multiple is
10xy(x + 3y) (x2+3xy+9y2)
(ii) Solution :
(x+4)2 (x-3)3, (x-1) (x+4) (x-3)2
By comparing (x+4) and (x+4)2, the highest term is (x+4)2.
By comparing (x-3)2 and (x-3)3, the highest term is (x-3)3
The extra term is (x-1).
So, the least common multiple is
(x-1)(x+4)2(x-3)3
The least common multiple is
(x-1)(x+4)2(x-3)3
(iii) Solution :
10 (9x2+6xy+y2) , 12 (3x2-5xy-2y2), 14 (6x4+2x3)
10 (9x2+6xy+y2) :
10 = 2 ⋅ 5
By factoring 9x2+6xy+y2, we get
9x2+6xy+y2 = 9x2+3xy+3xy+y2
= 3x(3x+y)+y(3x+y)
(9x2+6xy+y2) = (3x+y)(3x+y)
10 (9x2+6xy+y2) = 2 ⋅ 5 (9x2+6xy+y2) ----(1)
12(3x2-5xy-2y2) :
12 = 22⋅ 3
3x2-5xy-2y2 = (3x2-6xy+xy-2y2)
= 3x(x-2y)+y(x-2y)
= (3x+y) (x-2y) ----(2)
14(6x4+2x3) :
14 = 2 ⋅ 7
6x4+2x3 = 2x3(3x+1)
14(6x4+2x3) = 22 ⋅ 7⋅ x3 (3x+1) ----(3)
By comparing (1), (2) and (3), we get
= 22 ⋅ 5 ⋅ 7 ⋅ 3 ⋅ x³ ⋅ (3 x + y)²(3 x + 1)(x - 2y)
= 420 x3 (3 x + y)²(3 x + 1)(x - 2y)
So, the least common multiple is
420 x3 (3 x + y)2(3 x + 1)(x - 2y)
(iv) Solution :
3(a-1), 2(a - 1)2 , (a2-1)
= 3 (a- 1) -------(1)
2 (a - 1)2 = 2(a-1)(a-1) -------(2)
(a2-1) = (a+1) (a-1) -------(3)
By comparing (1), (2) and (3), we get
= 3 ⋅ 2 (a - 1)2 (a + 1)
So, the least common multiple is
6(a-1)2(a + 1)
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