C-04-Q028mediumsingle_mcq(am)n⋅(an)p(a^m)^n \cdot (a^n)^p(am)n⋅(an)p equals—aamnpa^{mnp}amnpbam+n+pa^{m+n+p}am+n+pব্যাখ্যাApplying power-of-a-power to each factor gives amn⋅anp=amn+np=an(m+p)a^{mn} \cdot a^{np} = a^{mn+np} = a^{n(m+p)}amn⋅anp=a. Factoring out the common matches option d.