$\text{ Let }{a}_1,a_2,a_3\dots \text{ be an A.P such that }\frac{a_1+a_2+\dots +a_p}{a_1+a_2+\dots +a_q}=\frac{p^3}{q^3},p\ne q\text{ then }\frac{a_6}{a_{21}}\text{ is }$

$\begin{array}{l}\text{take p=2,q=1 then the given equation becomes= }\frac{{\mathrm{a}}_1+{\mathrm{a}}_2}{{\mathrm{a}}_1}=8\Rightarrow \frac{{\mathrm{a}}_1+\left({\mathrm{a}}_1+\mathrm{d}\right)}{a_1}=8\\ \Rightarrow \mathrm{d}=6{\mathrm{a}}_1\\ \text{ Now }\frac{{\mathrm{a}}_6}{{\mathrm{a}}_{21}}=\frac{{\mathrm{a}}_1+5\mathrm{d}}{{\mathrm{a}}_1+20\mathrm{d}}=\frac{{\mathrm{a}}_1+5\times 6{\mathrm{a}}_1}{{\mathrm{a}}_1+20\times 6{\mathrm{a}}_1}=\frac{1+30}{1+120}=\frac{31}{121} \left(n\text{'}th term in A.P=a+\left(n-1\right)d\right)\end{array}$