Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$ Find all real numbers $x$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac76$$
I have tried to fiddle with it as follows: 
$$2^{3x} \cdot 6 +3^{3x} \cdot 6=12^x \cdot 7+18^x \cdot 7$$
$$ 3 \cdot 2^{3x+1}+ 2 \cdot 3^{3x+1}=7 \cdot 6^x(2^x+3^x)$$
Dividing both sides by $6$ gives us
$$2^{3x}+3^{3x}=7 \cdot 6^{x-1}(2^x+3^x)$$
Is this helpful? If so, how should I proceed form here? If not any hints would be greatly appreciated.

$$
\begin{array}{rcl}
6(8^x + 27^x) &=& 7(12^x + 18^x) \\
6(2^{3x} + 3^{3x}) &=& 7(3^x2^{2x} + 3^{2x}2^x) \\
\end{array}
$$
Substitute $a\!=\!2^x$ and $b\!=\!3^x$ for simplicity:
$$
\begin{array}{rcl}
6(a^3 + b^3) &=& 7(a^2b + ab^2) \\
6(a+b)(a^2 - ab + b^2) &=& 7ab(a+b) \\
6(a^2 - ab + b^2) &=& 7ab \\
6a^2 -13ab + 6b^2 &=& 0 \\
a^2 -\frac{13}{6}ab + b^2 &=& 0
\end{array}
$$
The left hand side can be factorized as $\left(a-\dfrac{3}{2}b\right)\left(a-\dfrac{2}{3}b\right)$.
$$
\left(a-\dfrac{3}{2}b\right)\left(a-\dfrac{2}{3}b\right)=0 \\
\begin{array}{rclcrcl}
2a &=& 3b &\text{or}& 3a &=& 2b \\
2.2^x &=& 3.3^x &\qquad\text{or}\qquad& 3.2^x &=& 2.3^x \\
x &=& -1 &\qquad\text{or}\qquad& x &=& +1
\end{array}
$$
Therefore, $x$ can be either $-1$ or $+1$.
$$ \boxed{x = \mp 1} $$
