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Given that $1<a<b$, how can you determine which is the larger, out of $a\sqrt[3]{b^2}$ and $b\sqrt[3]{a^2}$?

Thanks in advance.

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    $\begingroup$ Write $a = \sqrt[3]{a^3}$, $b = \sqrt[3]{b^3}$, and use properties of the cube root that you know. $\endgroup$ Sep 18, 2018 at 12:02
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    $\begingroup$ You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$
    – Shaun
    Sep 18, 2018 at 12:02
  • $\begingroup$ @Shaun will do next time. Thanks for the feedback. $\endgroup$
    – Sudera
    Sep 18, 2018 at 13:08

2 Answers 2

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HINT

We have

  • $x=a\sqrt[3]{b^2}\implies x^3=a^3b^2=a(ab)^2$
  • $y=b\sqrt[3]{a^2}\implies y^3=b^3a^2=b(ab)^2$

and recall that $f(x)=x^3$ is strictly increasing that is

$$x_1<x_2 \iff x_1^3<x_2^3$$

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It is $$a\sqrt[3]{b^2}<b\sqrt[3]{a^2}$$ if $$a^3b^2<b^3a^2$$ if $$a^2b^2(b-a)>0$$ and this is true.

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