Find $x$ such that $2^x+3^x-4^x+6^x-9^x=1$ The question:

Find values of $x$ such that $2^x+3^x-4^x+6^x-9^x=1$, $\forall x \in \mathbb R$.

Notice the numbers $4$, $6$ and $9$ can be expressed as powers of $2$ and/or $3$. Hence let $a = 2^x$ and $b=3^x$.
\begin{align}
1 & = 2^x+3^x-4^x+6^x-9^x \\
& = 2^x + 3^x - (2^2)^x + (2\cdot3)^x-(3^2)^x\\
& = 2^x + 3^x - (2^x)^2 + 2^x\cdot3^x-(3^x)^2 \\
& = a+b-a^2+ab-b^2 \\
0 & = a^2-ab+b^2-a-b+1
\end{align}
\begin{align}
0 & = a^2-ab+b^2-a-b+1 \\
& = 2a^2-2ab+2b^2-2a-2b+2 \\
& = (a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1) \\
& = (a-b)^2 + (a-1)^2 + (b-1)^2
\end{align}
This is where I am stuck. I am convinced that this factorisation could help solve the question, but I don't know how. Also, once we find values for $x$, we must prove that there are no further values of $x$. Could someone complete the question?

If the sum of a finite number of non-negative expressions is $0$, each of them has to be zero. 
In other words, when $a,b,c\geq0$
$a+b+c=0 \implies a=b=c=0$
You have done the hard part by showing that it can be written as the sum of three squares.
This means that $a=b$, $a=1$ and $b=1$. What does that tell you about $x$?
