<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] The equation $27ab+27=27a^3-b^3$ you have found may be written as
$$27a^3-27ab-27-b^3=0,$$
which may be viewed as a cubic in $b$. Its discriminant is
$$\Delta_b(a)=-4\cdot(-1)\cdot(-27a)^3-27\cdot(-1)^2=-(a^3+1)^2,$$
which is negative for all $a$ except for $a=-1$, when it is zero. This does not correspond to a real value of $x$, hence for all possible values of $a$ there is a unique real $b$ satisfying the polynomial, which is
$$b=3(a-1).$$
By definition we have $b=3^{\log_2a}$. So it remains to solve $3^{\log_2a}=3(a-1)$. By inspection we find the solutions $a=2$ and $a=4$ corresponding to $x=1$ and $x=2$. These are the only solutions, because such an exponential function intersects any line in at most two points.
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