# If $1<a<b$, which of the following is larger than the other: $a\sqrt[3]{b^2}$ and $b\sqrt[3]{a^2}$? [closed]

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

• Write $a = \sqrt[3]{a^3}$, $b = \sqrt[3]{b^3}$, and use properties of the cube root that you know. Sep 18, 2018 at 12:02
• 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. Sep 18, 2018 at 12:02
• @Shaun will do next time. Thanks for the feedback. Sep 18, 2018 at 13:08

• $x=a\sqrt[3]{b^2}\implies x^3=a^3b^2=a(ab)^2$
• $y=b\sqrt[3]{a^2}\implies y^3=b^3a^2=b(ab)^2$
and recall that $f(x)=x^3$ is strictly increasing that is
$$x_1<x_2 \iff x_1^3<x_2^3$$
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.