Results 1 to 3 of 3

Math Help - How are these 'the same'?

  1. #1
    Member
    Joined
    Aug 2008
    Posts
    182

    How are these 'the same'?

    Show that (\sqrt[3]{\frac{1}{16}})^{-1} + 8(\sqrt[3]{\frac{1}{16}})^2 = 3*2^\frac{1}{3}.

    Stuck again on this one.
    Any help would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    Yuck. I started with the left-hand side.

    (\sqrt[3]{\frac{1}{16}})^{-1} + 8(\sqrt[3]{\frac{1}{16}})^2 = 3*2^\frac{1}{3}

    The left hand side =((2^{-4})^{\frac{1}{3}})^{-1} + 2^3((2^{-4})^{\frac{1}{3}})^2

    Apply the (x^a)^b =x^{a \times b} and (x^a)(x^b)=x^{a+b} rules.

    =2^{(\frac{4}{3})}+2^{\frac{-8}{3}+3}

    =2^{\frac{4}{3}}+2^\frac{1}{3}}

    =2^{\frac{1}{3}}(2^1+1)

    =3(2^{\frac{1}{3}})

    I'm sure after my hours of work, someone will have already provided a much neater solution, but this is how I would approach the question. There is probably a much easier method, but this works nicely for me.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2008
    Posts
    182
    Thank you!
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum