Solve fo a step by step $$\frac{81^{a}+9^{a}+1}{9^{a}+3^{a}+1}=\frac{7}{9} \Rightarrow a = ? $$
 A: Hint: Let $x=3^a>0$. $$\frac{81^a+9^a+1}{9^a+3^a+1}=\frac{x^4+x^2+1}{x^2+x+1}=\frac{\left(x^2+1\right)^2-x^2}{x^2+x+1}$$
$$=\frac{\left(x^2+1+x\right)\left(x^2+1-x\right)}{x^2+x+1}=x^2+1-x=\frac{7}{9}$$
So you have a quadratic equation for $x$, and then $a=\log_3(x)$.
A: This equation reduces to:
$$ \frac{(3^a)^4 + (3^a)^2 + 1}{(3^a)^2 + 3^a + 1} = \frac{7}{9} $$
Let $3^a = x$
$$ \frac{x^4 + x^2 + 1}{x^2 + x +1} = \frac{7}{9}$$
$$ 9x^4 + 2x^2 - 7x + 2 = 0$$
Can you factorise the above equation and then find the values of a by substituting back in to $3^a = x$? Looks like a tricky factorisation.
Alternatively, you could use user236182's trick above to find the values of $x$.
A: Hint:
\begin{align}
\frac79&=\frac{81^a+9^a+1}{9^a+3^a+1}=\\
&=\frac{{(3^4)}^a+{(3^2)}^a+1}{{(3^2)}^a+3^a+1}=\\
&=\frac{{(3^a)}^4+{(3^a)}^2+1}{{(3^a)}^2+3^a+1};
\end{align}
Let $u=3^a$:
\begin{align}
\frac79&=\frac{u^4+u^2+1}{u^2+u+1}=\\
&=\frac{u^4+2u^2-2u^2+u^2+1}{u^2+u+1}=\\
&=\frac{{(u^2)}^2+2u^2+1+u^2-2u^2}{u^2+u+1}=\\
&=\frac{\left(u^2+1\right)^2-u^2}{u^2+u+1}=\\
&=\frac{\left(u^2-u+1\right)\left(u^2+u+1\right)}{u^2+u+1}=\\
&=u^2-u+1.
\end{align}
