The derivative of $f(x)$ is $f'(x)=3ax^{2}$. For $f(x)$ to be strictly increasing, $f'(x)>0$ for all $x$ in $(-1,1)$. This implies that $3ax^{2}>0$ for all $x$ in $(-1,1)$. Since $x^{2}$ is always non-negative, $a$ must be positive. Hence, $\mathrm{a}>0$. So,
Show more…