If $x, y, z\in\mathbb R^+ $ and $x^3+y^3=z^3,$ then prove that $x^2+y^2-z^2>6(z-x) (z-y). $ I made several unsuccessful attempts. Still couldn't think of a proper way to prove the inequality. Please suggest how to approach this problem. Thanks in advance.
EDIT 1. My approach (that I was talking about):
Given: $z^3=x^3+y^3.$
We have to prove:
$x^2+y^2-z^2>6(z-x) (z-y)$ i.e., $\underbrace{(z^2+zx+x^2) (z^2+zy+y^2) (x^2+y^2-z^2)}_{=E\text{ (say)}}>6(z^3-x^3) (z^3-y^3)=6x^3y^3.$
(Here one thing which I noticed is that $(x^2+y^2-z^2)>0,$ since each of the terms on the LHS except this one is positive and $6x^3y^3$ is also positive for $x, y, z>0.$)
Using AM $\ge$ GM, we have:
$E\ge 3zx\cdot3zy\cdot(x^2+y^2-z^2)\ge 9xyz^2(2xy-z^2).$
From here I couldn't think of a proper way to prove $E>6x^3y^3.$ But I'm still working on it. At present I'm trying to manipulate the expression $9xyz^2(2xy-z^2)$ to get the job done. If I find something useful I'll update here.

Because each expression is homogenous, we may assume that $ z = 1$ (by using the substitution $ x' = \frac{x}{z}$). The question becomes:

If $x^3 + y^3 = 1$, show that $x^2  - 6xy + y^2 + 6x  + 6 y - 7 > 0$.

Note: This 2-variable inequality type is common, and there are several ways of dealing with it by exploiting $x+y$.
Proof: Let $ w = x+y$. Since $ (x^3+y^3) < (x+y)^3 \leq 4(x^3 + y^3)$, thus $ 1 < w \leq \sqrt[3]{4} $.
$(x+y)(x^2  - 6xy + y^2 + 6x  + 6 y - 7)  \\ 
= x^3 +y^3  - 5xy (x+y) + 6(x+y)^2 - 7 (x+y) \\ 
= 1 - \frac{5}{3} ( (x+y)^3 - 1) + 6(x+y)^2 - 7 (x+y)   \\
= \frac{1}{3}(-5 w^3 + 18w^2 - 21w + 8) \\ 
= \frac{1}{3}(w-1)^2(-5w+8) 
$
Since $ 1 < w \leq \sqrt[3]{4} < \frac{8}{5}$, the final expression is  positive, and we conclude that $x^2  - 6xy + y^2 + 6x  + 6 y - 7 > 0$.

Of course, we didn't need to do the initial substitution, and could have shown that $$(x+y)\left[ x^2+y^2-z^2 - 6 ( z-y)(z-x)\right] = \frac{1}{3} ( x+y -z )^2 ( 8z - 5x - 5y )> 0.$$
However, it's hard to see how the equality holds by applying $(x+y)^3 = z^3 +  xy(x+y)$.
