Factorise using suitable identity 16a⁴+54ab³​

Question

Factorise using suitable identity 16a⁴+54ab³​

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Luna 2 weeks 2021-09-10T15:00:33+00:00 2 Answers 0 views 0

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    0
    2021-09-10T15:01:52+00:00

    16a^4+54ab^3

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    16a^4+54=2a(8a3+27)

    2a[(2a)3+(3)3)]

    a3+b3=(a+b)(a2+b2-ab)

    2a[(2a+3)(4a2+9-6a)]

    (2a) (2a+3)(4a2-6a+9)

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    hope this will help u

    0
    2021-09-10T15:02:14+00:00

    Step-by-step explanation:

    16a^4 + 54ab^3

    =2a(8a^3+27b^3)

    =2a{(2a)^3+(3b)^3}

    =2a(2a+3b)(4a^2-6ab+9b^2)

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