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Factorise each of the following:(iv) 64a3–27b3–144a2b+108ab2 Q.8(4)

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What is the best way for solving the question from class 9th ncert math of Polynomials chapter of Ncert of exercise 2.5 of math. What is the best way for solving this question please guide me the best way for solving this question Factorise each of the following:(iv) 64a3–27b3–144a2b+108ab2

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  1. The expression, 64a3–27b3–144a2b+108ab2can be written as (4a)3–(3b)3–3(4a)2(3b)+3(4a)(3b)2

    64a3–27b3–144a2b+108ab2=
    (4a)3–(3b)3–3(4a)2(3b)+3(4a)(3b)2

    =(4a–3b)3

    =(4a–3b)(4a–3b)(4a–3b)

    Here, the identity, (x – y)3 = x3 – y3 – 3xy(x – y) is used.

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