Can i perform this step? a,b $\in$ R
$${5^{a-b}}\cdot{27^{-(a+b)}}={5^1}\cdot {27^1} \Rightarrow a^{2}-b^{2}=?$$

My Solution:
$a-b=1$
$a+b=-1$  then $a=0$ ; $b=-1$
Since $a$ and $b$ are real numbers, I believe, I cannot perform above solution. Am I right?
Note that for $(a,b)=(0,-1)$ we have 
$$5^{a-b}\cdot27^{-(a+b)}=5^1\cdot27^1,$$
and $a^2-b^2=-1$, and that for $(a,b)=\left(\frac{\log(5^1\cdot27^1)}{\log(5/27)},0\right)$ we have
$$5^{a-b}\cdot27^{-(a+b)}=5^a\cdot27^{-a}=\left(\frac{5}{27}\right)^{\frac{\log(5^1\cdot27^1)}{\log(5/27)}}=5^1\cdot27^1,$$
and $a^2-b^2=\left(\frac{\log(5^1\cdot27^1)}{\log(5/27)}\right)^2\geq0$. So the value of $a^2-b^2$ is not uniquely determined by the given equation. In particular, this shows that the step you took is not valid.
In the above equation a and b are not whole numbers but real numbers
$a-b=1$

$a+b=-1$

We know that
$a^2-b^2=(a+b)(a-b)$

$a^2-b^2=(-1)(1)$
$a^2-b^2=-1$