Let, #D=|(1,a,a^2,p),(1,b,b^2,q),(1,c,c^2,r),(1,d,d^2,s)|#,
where, #p=a^3+bcd, q=b^2+cda, r=c^3+dab, s=d^3+abc#.
Note that, #D# can be split as the sum of two determinants
#D_1 and D_2#, where,
#D_1=|(1,a,a^2,a^3),(1,b,b^2,b^3),(1,c,c^2,c^3),(1,d,d^2,d^3)|,# &,
#D_2=|(1,a,a^2,bcd),(1,b,b^2,cda),(1,c,c^2,dab),(1,d,d^2,abc)|.#
So, #D=D_1+D_2#.
By #R_1xxa, R_2xxb, R_3xxc, R_4xxd,# we have,
#(abcd)D_2=|(a,a^2,a^3,abcd),(b,b^2,b^3,bcda),(c,c^2,c^3,cdab),(d,d^2,d^3,dabc)|, or,#
#D_2=|(a,a^2,a^3,1),(b,b^2,b^3,1),(c,c^2,c^3,1),(d,d^2,d^3,1)|.#
Next, the interchange between #C_1" & "C_4# will give #-D_2,#
#i.e., -D_2=|(1,a^2,a^3,a),(1,b^2,b^3,b),(1,c^2,c^3,c),(1,d^2,d^3,d)|.#
Similarly, #C_2harr C_4# will yield #-(-D_2)=D_2#.
#:.D_2=|(1,a,a^3,a^2),(1,b,b^3,b^2),(1,c,c^3,c^2),(1,d,d^3,d^2)|.#
Finally, #C_3harrC_4# will result in #-D_2#.
#:.-D_2=|(1,a,a^2,a^3),(1,b,b^2,b^3),(1,c,c^2,c^3),(1,d,d^2,d^3)|.#
But then, this means that, #-D_2=D_1#, or,
#D_1+D_2=D=0#, as Respected Maganbhai P. has readily
shown!
#color(blue)("Enjoy Maths.!")#