What is bigger, $\sqrt[3]{3}+\sqrt[3]{24}$ or $\sqrt[3]{2}+\sqrt[3]{54}\;$?

Question

I have a math problem involving radicals. I tried to do it and I didn't find how to solve it.

What is bigger, $\sqrt[3]{3}+\sqrt[3]{24}$ or $\sqrt[3]{2}+\sqrt[3]{54}\;$?

The answer can easily be found using a calculator, but I want a analytic solution.

Thanks in advance.

Answer 1

We wish to compare $a:=\sqrt[3]{3}+\sqrt[3]{24}$ and $b:=\sqrt[3]{2}+\sqrt[3]{54}$. We have $$a=\sqrt[3]{3}+2\sqrt[3]{3}=3\sqrt[3]{3}\quad\text{and}\quad b=\sqrt[3]2+3\sqrt[3]{2}=4\sqrt[3]2.$$ Now $a^3=81$ and $b^3=128$. Hence $a<b$.

Answer 2

Let, $$x=\sqrt[3]{3}+\sqrt[3]{24}$$ $$y=\sqrt[3]{2}+\sqrt[3]{54}$$ Our aim is to show that $y>x$. Taking, $\sqrt[3]{3}$ common in x, $$x=\sqrt[3]{3}(1+\sqrt[3]{8})=\sqrt[3]{3}(1+2)$$ $$\sqrt[3]{3}(3)=3^{\frac{1}{3}+1}=3^{\frac{4}{3}}$$ Taking $\sqrt[3]{2}$ common in y, $$y=\sqrt[3]{2}(1+\sqrt[3]{27})=\sqrt[3]{2}(1+3)= \sqrt[3]{2}(2^{2})=2^{\frac{1}{3}+2}=2^{\frac{7}{3}}$$ $$x=3^{\frac{4}{3}} y=2^{\frac{7}{3}}$$ $$x^{3}=3^{4} , y^{3}=2^{7}$$ $$\implies x^{3}<y^{3}$$ $$x<y$$ Hence proved.

Answer 3

$$\sqrt[3]{3}+\sqrt[3]{24}<\sqrt[3]{2}+\sqrt[3]{54}$$

Proof:-

$\sqrt[3]{3}+\sqrt[3]{24}<\sqrt[3]{2}+\sqrt[3]{56}$

$(\sqrt[3]{3}(1+\sqrt[3]{8}))^3<(\sqrt[3]{2}+\sqrt[3]{54})^3$

$81<56+3\sqrt[3]{216}+3\sqrt[3]{5832}$

$25<3×6×3×18$

$25<72$

Q.E.D