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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10

Find the L.C.M. of the given expressions (2x2 – 3xy)2, (4x – 6y)3, (8x3 – 27y3) - Mathematics

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Question

Find the L.C.M. of the given expressions

(2x2 – 3xy)2, (4x – 6y)3, (8x3 – 27y3)

Sum

Solution

(2x2 – 3xy)2 = x2 (2x – 3y)2

(4x – 6y)3 = 23 (2x – 3y)3

= 8 (2x – 3y)3

(8x3 – 27y3) = (2x)3 – (3y)3

= (2x – 3y) [(2x)2 + 2x × 3y + (3y2)] ...[using a3 – b3 = (a – b) (a2 + ab + b2)]

(2x – 3y) (4x2 + 6xy + 9y2)

L.C.M. = 8x2 (2x – 3y)3 (4x2 + 6xy + 9y)2

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GCD and LCM of Polynomials
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Chapter 3: Algebra - Exercise 3.2 [Page 97]

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Samacheer Kalvi Mathematics [English] Class 10 SSLC TN Board
Chapter 3 Algebra
Exercise 3.2 | Q 2. (vi) | Page 97
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