Show that $0 = 2a^3-5ab^2+25b^3$ has no other integer solutions than $a = b = 0$. I am trying to solve the following problem:
I have the equation $0 = 2a^3-5ab^2+25b^3$, where $a,b \in \mathbb Z$. Obviously, $a = b = 0$ is a solution of this equation. But how can I show that there are no other integer solutions? 
I considered the function $h : \mathbb R^2 \to \mathbb R$, defined by $h(x,y) = 2x^3-5xy^2+25x^3$. I found that the only extreme value is at $x = y = 0$ and $h(0,0) = 0$. But I cannot conclude that there are not other integer roots, because not every root must be necessarily an extremum.
Maybe there is a better approach without Analysis. Thanks.
 A: Suppose $b \not= 0$. Then, dividing throughout by $b^3$, we have
$$2\left(\frac{a}{b}\right)^3 - 5\left(\frac{a}{b}\right) + 25 = 0$$
By the rational root theorem, since $\frac{a}{b}$ is rational, it is necessary that $a' | 25$ and $b'|2$, where $a' = \frac{a}{d}, b' = \frac{b}{d}, (a,b) = d$. From this, we have $a' = \pm 1, \pm 5, \pm 25$, and $b' = \pm 1, \pm 2$. We test all possible value of $\frac{a}{b}$ to conclude that there are no solutions if $b\not=0$.
Hence, $b = 0$ and it follows that $a = 0$.
You could do the same thing by dividing throughout by $a^3$ instead as well, using the same argument and approach.
A: Hint: Divide with $b^3$ and set $x=\frac{a}{b}$. Then by the rational root theorem the polynomial $$0=2x^3-5x+25$$ has no rational roots (case $b=0$ is excluded).
A: Hint: Let $f(a,b)=2a^3-5ab^2+25b^3$. Now consider, $a\ne0$.  
So, $f(a,b)=a^3\bigg(2-5\bigg(\dfrac{b}{a}\bigg)^2+25\bigg(\dfrac{b}{a}\bigg)^3\bigg)$. Consider $\dfrac{b}{a}=x$.  
So, show that $g(x)=2-5x^2+25x^3$ has no integer solution.
Again, consider $b\ne 0$ as repeat the process.
