Given xyz>0, possible signs for x, y, z will be as follows:
x y z
+ + +
+ - -
- + -
- - +
1. y<0
Looking at the above list if y<0 then either x is positive and z is negative or x is negative and z is positive.
In either case x * (y^2) * (z^3) <0
So. 1 is sufficient.
2. x>0
Again from the table above is x>0, then either both y and z are positive or negative.
When both y and z are positive x * (y^2) * (z^3)>0
When both y and z are negative x * (y^2) * (z^3)<0
Therefore insufficient.
Hence answer is A
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