If $$p^{2}+q^{2}-29=2pq-20=52-2pq$$, then the difference between the maximum and minimum possible value of $$(p^{3}-q^{3})$$

Given that 2pq - 20 = 52 - 2pq => 4pq = 72 => pq = 18 ----(1)

Now, $$p^2+q^2-29=2pq-20$$ => $$p^2+q^2-2pq=9$$ => $$\left(p-q\right)^2=9$$ => $$p-q=\pm\ 3$$

Also, $$p^2+q^2-29=2pq-20$$ => $$p^2+q^2=2pq+9=2\left(18\right)+9=45$$

Now, $$p^3-q^3=\left(p-q\right)\left(p^2+pq+q^2\right)$$ = $$\left(p-q\right)\left(45+18\right)$$ = $$\left(p-q\right)\left(63\right)$$

=> When p-q = -3 => The value is 63(-3) = -189 and when p-q = 3 => The value is 63(3) = 189.

=> The difference = 189 - (-189) = 378.