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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is -3*x^3 <= -3?

1) -3x<3
2) x <0

When you modify the condition and the question, x^2+x+1>0 is always right from -3x^3<=-3?, x^3>=1?, (x-1)(x^2+x+1)>=0?. It becomes x-1>=0?, x>=1?. So, there is 1 variable, which should match with the number of equations. You need 1 more equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
In 1), from x>1, x=0 is no and x=3 is yes, which is not sufficient.
In 2), when x<0, it is always no, which is sufficient. Therefore, the answer is B.


-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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-3\(x^3\)<=-3

Rearrange the above equation

3\(x^3\)-3>=o

3(\(x^3\)-1)>=0

(\(x^3\)-1)>=0

\(x^3\)>=1 OR x>=1

Statement 1: -3x<3

-3x-3<0

Multiply the equation with -1 and inverse the sign

3x+3>0

3(x+1)>0

So X can be 0 or >1 [Insuff]

Statement 2: x<0...............Directly gives the answer no.[Suff]

Answer: B.
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Is \(-3x^3 <= -3?\)
= \(3x^3 >= 3?\)
= \(3x^3 - 3>= 0?\)
= \(3(x^3 - 1) >= 0?\)
\(x^3 >= 1\)
\(x >= 1\)?

1) \(-3x<3\)
\(-3x -3 < 0\)
\(-3(x+1) < 0\)
\(x > -1\)

Insufficient.

2) \(x < 0\)

Sufficient.

Answer is B.
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