Solve fo a step by step $$\frac{81^{a}+9^{a}+1}{9^{a}+3^{a}+1}=\frac{7}{9} \Rightarrow a = ? $$

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 $x$.
