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    (Original post by Zacken)
    No worries, I can see why you'd be confused though, don't worry about it!
    Lol yeah, that actually makes the qn a lot more difficult. Don't think I recall doing a question like that before.
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    (Original post by Zacken)
    Simples: x^{2/3} - 2x = 0 \Rightarrow x(x^{-1/3} - 2) = 0 so either x=0 or x^{-1/3} = 2 the latter is easy to solve, since x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}.
    Yeaaaah fully get it now!
    Thankyou so much
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    (Original post by Ggffffhh)
    Yeaaaah fully get it now!
    Thankyou so much
    No worries, glad that helped.
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    (Original post by Ggffffhh)
    Yeaaaah fully get it now!
    Thankyou so much
    (Original post by TheTopStudent)
    Have to agree with Bath Student, there is much simpler way to solve this not involving the reciprocal and third power, just confusing OP otherwise.
    Yep, defo confused the OP. I don't really see the difference - they are logically equivalent, it's just applying the inverse at a later time - mine allows you to observe the solution from a straightforward injectivity argument which is why I prefer it.

    But hey, why don't you post your 'alternative' method for others to benefit from!
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    (Original post by Zacken)
    Yep, defo confused the OP. I don't really see the difference - they are logically equivalent, it's just applying the inverse at a later time - mine allows you to observe the solution from a straightforward injectivity argument which is why I prefer it.

    But hey, why don't you post your 'alternative' method for others to benefit from!
    Cos I already did. Those smiley faces are painfully obnoxious..
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    (Original post by Bath~Student)
    Cos I already did. Those smiley faces are painfully obnoxious..
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    (Original post by Zacken)
    Simples: x^{2/3} - 2x = 0 \Rightarrow x(x^{-1/3} - 2) = 0 so either x=0 or x^{-1/3} = 2 the latter is easy to solve, since x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}.
    Could you explain
    x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}

    and how x^-1/3 = 2 = 8^1/3?
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    (Original post by hellomynameisr)
    Could you explain
    x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}

    and how x^-1/3 = 2 = 8^1/3?
    2 is the same thing as 8^{1/3}. So if I say that x^{-1/3} = 2 then it must mean that x = 8^{-1}.
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    (Original post by hellomynameisr)
    Could you explain
    x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}

    and how x^-1/3 = 2 = 8^1/3?
    It isn't. It is, however, a stylistic abomination.
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    (Original post by Zacken)
    2 is the same thing as 8^{1/3}. So if I say that x^{-1/3} = 2 then it must mean that x = 8^{-1}.
    Ohhh I see. Thank you
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    (Original post by hellomynameisr)
    Ohhh I see. Thank you
    No worries.
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    I multiplied both sides by 3 which gave me 3x^2=18x, then divided both sides by x which gave me 3x=18 and then solved for x which gave me x=6
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    (Original post by celiacrowther)
    I multiplied both sides by 3 which gave me 3x^2=18x, then divided both sides by x which gave me 3x=18 and then solved for x which gave me x=6
    The question is 3x^(2/3) not (3x^2)/3.
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    (Original post by Zacken)
    Simples: x^{2/3} - 2x = 0 \Rightarrow x(x^{-1/3} - 2) = 0 so either x=0 or x^{-1/3} = 2 the latter is easy to solve, since x^{-1/3} = 2 = 8^{1/3} = \left(\frac{1}{8}\right)^{-1/3}.
    It's obviously personal preference but I like your way the best.

    A method without dividing the equation or cubing both sides feels nicer to me. Factorising is underrated.
 
 
 
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