Number of real solutions of the equation $1+8^x+27^x = 2^x+12^x+9^x$ 
Find the number of real solutions $x\in\mathbb{R}$ of the equation
  $$
1+8^x+27^x = 2^x+12^x+9^x
$$

My Attempt:
Let $2^x=a>0$ and $3^x=b>0$ where $x\in \mathbb{R}$. This allows us to change the equation to
$$
1+a^3+b^3 = a+a^2b+b^2
$$
This can be rewritten as
$$
(a+b)^3-3ab(a+b)+1 = a+ab(a+b)
$$
How can I solve the problem from this point?
 A: Hint: by rearrangement inequality
$$a^3+b^3+c^3 = a^2b+b^2c+c^2a$$
happens iff $a=b=c$. 
A: A different method. Define:
$$f(a,b)= 1 + a^3 + b^3 - (a + a^2 b + b^2)$$
Show by solving $f'_a = f'_b=0$ that the minimum of $f$ is at $a=b=1$, and that $f(1,1)=0$, proving there is a single solution, $x=0$. 
A: If the equation $f(a,b,c)=0$ is symmetrical in the sense that $f(a,b,c)=f(\overline{a,b,c})$ where  $\overline{a,b,c}$ represents any permutation of $\{a,b,c\}$ then if it has an extreme point it must be $a=b=c$. 
Proof is very simple. Take that it has an extreme point for $a_{0}>b_{0}>c_{0}$. Then by permutation it would have more extreme points where $b_{0}>a_{0}>c_{0}$ or $a_{0}>c_{0}>b_{0}$. Because of this it must be only $a_{0}=b_{0}=c_{0}$.
It remains to prove that the extreme exists and that it is a minimum in this particular case.
For any $a=b=c$ we have $f(a,b,c)=a^{3}+b^{3}+c^{3}-a^{2}b-b^{2}c-c^{2}a=0$.
Let us have a point $a_{1} \geq b_{1} \geq c_{1} \geq 0$ where it is not 
$a_{1} = b_{1} = c_{1}$ 
It would be $(a_{1}^{3}-a_{1}^{2}b_{1})+(b_{1}^{3}-b_{1}^{2}c_{1})+(c_{1}^{3}-c_{1}^{2}a_{1})>0$ because each $a_{1}^{3}-a_{1}^{2}b_{1} \geq 0,b_{1}^{3}-b_{1}^{2}c_{1} \geq 0, c_{1}^{3}-c_{1}^{2}a_{1} \geq 0$ but at least one is different from $0$.
So $0$ is a minimum. The solution to the above equation follows.
