consider negative range to find extreme scenarios
A) p^2 > q^2
Insufficient as
q can be 2 while p can be -3 still satisfying equation 9> 4 but -3 < 2
also q can be 2 and p can be 3 satisfying equation 9> 4 also 3> 2
Squares hide the original sign
B) if p =-2 and q= -3
p^3 = -8
q^3 = -27
-8 > -27 and -2 > -3 therefore p > q
Also, p=2 , q=-3 then 8> -27 therefor p>q
similarly if we choose one + one -ve to satisfy the equation, we will get p>q
similar behavior
...
A) p^2 > q^2
Insufficient as
q can be 2 while p can be -3 still satisfying equation 9> 4 but -3 < 2
also q can be 2 and p can be 3 satisfying equation 9> 4 also 3> 2
Squares hide the original sign
B) if p =-2 and q= -3
p^3 = -8
q^3 = -27
-8 > -27 and -2 > -3 therefore p > q
Also, p=2 , q=-3 then 8> -27 therefor p>q
similarly if we choose one + one -ve to satisfy the equation, we will get p>q
similar behavior
...





