# Inequality involving radicals and using the factorization of x^3-y^3

My attempt at a solution :

Okay, so I can get the solution by working backwards ( working out each side separately, cubing it, finding the final expression, and then I can just work my way back up to show that the original inequality was true) , but this is somewhat messy, and the question clearly asks me to use the factorization of $x^3 -y^3$, which I see no point to apply in my solution.

Could someone point me towards how to add this in my answer?

## 1 Answer

Hint:

Let $x^3 = a^3$ and $y^3= a^3-h$. Note that $x^3-y^3=h$. Factorize $x^3-y^3$ and use the given constraints on $h$ to prove the inequalities. $$h=x^3-y^3=(x-y)(x^2+xy+y^2)=(a-\sqrt[3]{a^3-h})(a^2+a(a^3-h)^{1/3}+(a^3-h)^{2/3}).$$

• Sorry, I should've clarified that I only want a hint. Not much point to it when I'm given the whole answer =) I appreciate the effort, but could you edit it to make it only the first line? Thanks. – user440261 Sep 9 '17 at 3:52
• Also, could you tell me how exactly you came up with the first part? Was it just trying plausible values of x^3 and y^3 till you got it? – user440261 Sep 9 '17 at 12:20
• @Saad The idea is to eliminate the cube-root sign from $\sqrt[3]{a^3-h}$ by taking cubes. – Math Lover Sep 9 '17 at 12:26
• Ahh. That makes sense. Thank you. – user440261 Sep 9 '17 at 15:28