Help solve this question urgently

325. If $\underset{x\to 1}{\lim }\frac{x^4-1}{x-1}=\underset{x\to 1(-m)}{\lim }\frac{x^3+m^3}{x^5+m^5}$, then the value of m ² is
(a) $\frac{1}{3}$
(b) $\frac{2}{5}$
(c) $\frac{3}{20}$
(d) $\frac{5}{24}$