Yes, if $a,b \neq 0$ and are real. We have $a-b = \frac{(a-b)(a^2+ab+b^2)}{ab}$. Since $a,b \neq 0$, we can clear the denominator and move to RHS then factor as $(a-b)(a^2+b^2) = 0$. But since $a,b$ are nonzero, we must have $a-b = 0 \implies a=b$