How do I factorise $729^x-243^x-81^x+9^x+3^x-1$?

Question

How do I factorise $$729^x-243^x-81^x+9^x+3^x-1 ?$$

I know the answer is $(3^x-1)(9^x-1)(27^x-1)$, but I have no idea how to go about it. This is actually a step of another question in my book. Would someone please help?

Answer 1

Hint Write the expression in terms of $u := 3^x$. For example, $9^x = (3^2)^x = (3^x)^2 = u^2$.

Answer 2

From $y=3^x$ we get:

$$y^6-y^5-y^4+y^2+y-1=0$$ Can you see $y=1$ and $y=-1$ are solutions to this?

Hence we have: $$y^6-y^5-y^4+y^2+y-1=(y^2-1)(ay^4+by^3+cy^2+dy+f)$$ Work out the quartic if then try to reduce it further.