If ax² + bx³ = 5, where a and b are non-zero numbes, what is the value of a + b?
(1) ax² = a
(2) bx³ = b

The possible value of X is +/-1 from the first condition. Therefore if we proceed ahead with this without leaving it here, there will be 2 equations for a and b : a + b equal to 5 and a – b equal to 5. Solving both these equations, we get the value of a + b equal to 5. So even option A is sufficient.

The important thing to remember here is that, when a number is squared, it will always result in a positive, regardless of whether the original number was negative or positive. This is not the case with a number that is cubed. A cube will only be positive if the original integer was positive; otherwise, it will be negative. That having been said, statement (1) is insufficient. While we know that X can be nothing other than 1 or –1 in order for ax^2to equal a, we are not told specifically if X is 1 or –1. Statement (2), however, is sufficient. Because X is cubed, we know that it had to be 1 from the beginning. Statement (2) tells us that a + b = 5

ax² = a
x² = 1
x = 1 or x = -1

If x = 1, then
ax² + bx³ = a × 1² + b × 1³ = a + b = 5

If x = 1, then
ax² + bx³ = a × (-1)² + b × (-1)³ = a – b = 5
But we don't know what a + b equals to in this case (if x = 1).

Therefore statement (1) by itself is NOT sufficient.