<s>[INST] Find all real solutions of $6^x+1=8^x-27^{x-1}$ 
Find all real solutions of $6^x+1=8^x-27^{x-1}$.

Things I tried:
We want solutions of
$$2^x3^x+1 = (2^x)^3-\frac{(3^x)^3}{27}.$$ Write $a=2^x$ and $b=3^x$. This gives
$$ab+1 = a^3-\frac{b^3}{27}$$
or
$$27ab+27=27a^3-b^3$$
How to continue?
 [/INST] As you did, let $a=2^x, b=3^x$. We have $$ab+1=a^3-\frac{b^3}{27}$$
Note the standard identity $$u^3+v^3+w^3-3uvw=(u+v+w)\left(\frac{(u-v)^2+(v-w)^2+(w-u)^2}{2}\right)$$
Take $u=a, v=-\frac{b}{3}, w=-1$. Then $u^3+v^3+w^3-3uvw=a^3-\frac{b^3}{27}-1-ab=0$.
Thus $u+v+w=0$ or $(u-v)^2+(v-w)^2+(w-u)^2=0$. In the latter case, we get $u=v=w$ so $a=-\frac{b}{3}=-1$, so $a=-1, b=3$. But clearly $2^x=a \not =-1$, a contradiction.
Thus $u+v+w=0$.
Thus $a-\frac{b}{3}-1=0$, i.e. $2^x=3^{x-1}+1$.
Write $x=y+1$, and rewrite as $1^y+3^y=2^y+2^y$. Finally use the theorem I prove here. (Quoted below)

Theorem: Let $a, b, c, d$ be real numbers such that $0<a<b \leq c<d$. Then the equation $$a^x+d^x=b^x+c^x$$ has 
  
  
*
  
*Exactly two solutions, $x=0$ and $x=t>0$ for some $t$, if $ad-bc<0$
  
*Exactly two solutions, $x=0$ and $x=t<0$ for some $t$, if $ad-bc>0$
  
*Exactly one solution, $x=0$, if $ad-bc=0$
  

with $(a, b, c, d)=(1, 2, 2, 3)$, $ad-bc=-1<0$, and $y$ in place of $x$.
Thus there are exactly two solutions for $y$, given by $y=0$ and $y=t>0$, some $t$. By inspection, $y=1$ is a solution. Thus all real solutions for $y$ are $y=0, 1$.
Thus all real solutions for $x$ are $x=1, 2$.
</s>