${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ - 3/2}} = $

  • A

    ${(x/4)^3}$

  • B

    ${(4x)^3}$

  • C

    $8{x^3}$

  • D

    None of these

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