This is a tricky one.

(1) Gives \(x \leq -3\) as x is negative. Sufficient

(2)

=> \(x < \sqrt[3]{-9}\)

=> \(x < - \sqrt[3]{9}\)

We know -8 is the nearest perfect cube and -27 is next nearest

so

\(-\sqrt[3]{27} < -\sqrt[3]{9} < -\sqrt[3]{8}\)

\(-3 < -\sqrt[3]{9} < -2\)

\(-\sqrt[3]{9}\) is a point between -3 and -2 on the number line and we got [highlight]\(x < - \sqrt[3]{9}\)[/highlight]

So this doesn't answer our question if \(x \leq -3\) as there can be some values of x that can be \(x \geq -3\) using statement 2 rephrase that we got above.

In sufficient

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