Solve the question completly

Solution 
LHS,
(a^p+q )² = a^2p+2q     {because (a^m)ⁿ = a^mn}
(a^q+r)² = a^2q+2r
(a^r+p)² = a^2r+2p
(a^p . a^q . a^r)⁴ = (a^p+q+r)⁴     {because a^m  . aⁿ = a^m+n}
                      = a^4p+4q+4r
Now, putting the values in LHS, we have,
$$\begin{gathered} =\frac{a^{2p+2q}.a^{2q+2r}.a^{2r+2p}}{a^{4p+4q+4r}} \{because a^m.a^n=a^{m+n}\} \\ =\frac{a^{2p+2q+2q+2r+2r+2p}}{a^{4p+4q+4r}} \\ =\frac{a^{4p+4q+4r}}{a^{4p+4q+4r}} \\ =1 \\ =RHS \\ Hence Proved. \end{gathered}$$