given: x, y are integers

considering x^3 = -125
-> x = -5
x^7 = (-5)^7 = -ve number
6^y where y is any integer
take y = -2, 6^y is positive
y = 0, 6^y is positive
y = 2, 6^y is positive
so, x^3 is less than 6^y.
it is sufficient to answer.

consider y^2 = 36,
-> y = +6 or -6
we don`t have any information of x^7. so, it is not sufficient.

Hence, it is A alone is sufficient

CSASHISHPANDAY wrote:if X & Y are integer is X^7 is greater that 6^Y

1. X^3=-125
2. Y^2=36

Target question: Is x^7 > 6^y?

Statement 1: x^3 = -125
There's a nice rule that says, "An odd exponent preserves the sign of the base"
In other words, (any negative number)^(odd integer) = negative number
. . . and (any positive number)^(odd integer) = positive number

So, statement 1 essentially tells us that x must be a negative number. Now, we could go further and determine that x = -5, but that's not really necessary here. Knowing that x is negative provides enough information to answer our target question with certainty.

First, since x is negative, we know that x^7 must be negative.
Second, we know 6^(any value) will equal a positive number.

So, it must be the case that Is x^7 is not greater than 6^y?
Statement 1 is SUFFICIENT

Statement 2: y^2 = 36
From this, we can conclude that y = 6 or y = -6.
Let's examine both cases:
case a: y = 6.
In this case, we cannot determine whether x^7 is greater than 6^y, since we have no idea what x equals.
case b: y = -6.
In this case, we cannot determine whether x^7 is greater than 6^y, since we have no idea what x equals.
Statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent