Question

i tried this sum so many times and posted it also so many times, but didn;t get accurate answer

Answer 1

it just amounts to alot of subbing and simplifying

Answer 2

if you can determine x^3; then y^3 and z^3 are the same construct but with the appropriate variable combinations

Answer 3

(n^2 - pq)^3

1 3 3 1
n^2^3 n^2^2 n^2^1 n^2^0
-pq^0 -pq^1 -pq^2 -pq^3
-----------------------------
n^6 -3n^4pq +3(npq)^2 -(pq)^3

Answer 4

its that xyz part that might get hairy