MHB Solving Inequality Problem: Proving Radical Expressions with Cube Roots

[anemone]

Prove $\sqrt[3]{1−12\sqrt[3]{65^2} + 48\sqrt[3]{65}} -\sqrt[3]{63}\gt \sqrt[3]{1−48\sqrt[3]{63} + 36\sqrt[3]{147}} -4 $

 

[kaliprasad]

Prove $\sqrt[3]{1−12\sqrt[3]{65^2} + 48\sqrt[3]{65}} -\sqrt[3]{63}\gt \sqrt[3]{1−48\sqrt[3]{63} + 36\sqrt[3]{147}} -4 $

Spoiler

first let us simplify LHS
$\sqrt[3]{1−12\sqrt[3]{65^2} + 48\sqrt[3]{65}}$
this appears to be a cube whose cube root is of the form
$a\sqrt[3]{65}-b$
cube it to get
$65a^3 - 3a^2\sqrt[3]{65^2}b+3ab^2\sqrt[3]{65}-b^3 = 1−12\sqrt[3]{65^2} + 48\sqrt[3]{65}$
compare both sides to get
$3a^2b= 12$
$3ab^2 = 48$
giving a = 1 and b= 4
but we need to check the rational part
65* 1^3 - 4^3 = 1 so it is correct and hence

LHS becomes

$\sqrt[3]{1−12\sqrt[3]{65^2} + 48\sqrt[3]{65}}- \sqrt[3]{63} = \sqrt[3]{65}-4 - \sqrt[3]{63}\cdots(1)$

similarly trying the RHS with $a\sqrt[3]{7}-b\sqrt[3]{3}$ and failing and the trying with $a- \sqrt[3]{63}$
we get the RHS as
$\sqrt[3]{1−48\sqrt[3]{63} + 36\sqrt[3]{147}} -4=4- \sqrt[3]{63}- 4\cdots(2)$

hence from (1) and(2) we get the result

 

[anemone]

Well done kaliprasad.

My solution:

Spoiler

Rewrite the inequality as $∛(1−12∛65² + 48∛65)+4\gt ∛(1−48∛63 + 36∛147)+∛63$.

If we can prove LHS is greater than $4$ and RHS equals $4$, then we are done.

First, note that $1−12∛65² + 48∛65=∛65³-3(∛65)(∛64)(∛65-∛64)-∛64³=(∛65-∛64)^3>0$, so it must be true that $LHS=∛(1−12∛65² + 48∛65)+4>4$.

Next, note that $∛(1−48∛63 + 36∛147)=∛(∛64³-3(∛64)(∛63)(∛64-∛63)-∛63³)=∛(∛64-∛63)³=∛64-∛63=4-∛63$, that suggests $RHS=∛(1−48∛63 + 36∛147) + ∛63=4$, this completes the proof.