Let $a,b,c$ be positive integers such that $a^3+b^3=2^c.$ Show that $a=b$. 
Let $a,b,c$ be positive integers such that $$a^3+b^3=2^c.$$ Show that $a=b$.

I have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=2^c =2^x\cdot2^y$$ now it can only be that $a$ and $b$ are both odd or even since they sum to an even number. Thus if $a$ and $b$ are both odd I have that $a^2$ is odd, $ab$ is odd and $b^2$ is odd. This would imply that $a^2-ab+b^2 = 2^y =1$ which in turn implies that $a^3+b^3 = a+b \implies a=b.$ The problem I have is that if I would have considered that both $a$ and $b$ are even I would have gotten that $a=2t, b=2k$ from where $$8t^3+8k^3=2^c \implies t^3+k^3=2^{c-3}$$ but I couldn't deduce anything from here why cannot $a$ and $b$ be even?

Write $a = 2^x \cdot m$ and $b = 2^y \cdot n$ where $m$ and $n$ are odd. WLOG $x \geq y$. Now,
$$2^{3x} \cdot m^3 + 2^{3y} \cdot n^3 = 2^{3y} \cdot (2^{3x-3y} \cdot m^3 + n^3) = 2^c$$ If $x > y$, we have a contradiction since in that case an odd number divides LHS. So, $x = y$ and so you would arrive at your argument.
