Question

What values of $x$ satisfy $\sqrt[3]{x} + x^{1/3} - 2 \leq 0$ ($x$ is a real number)?

• A. $-8 \leq x \leq 1$
• B. $-1 \leq x \leq 8$
• C. $1<x<8$
• D. $1 \leq x \leq 8$
• E. $-8 \leq x \leq 8$

Hint:

Substitute cube roots with a new variable to simplify solving inequalities.

Answer 1

The Correct Option is A

Let $t = \sqrt[3]{x}$. Then the inequality becomes: \[ t + t^3 - 2 \leq 0 \] \[ t^3 + t - 2 \leq 0 \] Factorizing: \[ t^3 + t - 2 = (t - 1)(t^2 + t + 2) \] The quadratic $t^2 + t + 2>0$ for all real $t$. Thus inequality reduces to: \[ (t - 1) \leq 0 \quad \Rightarrow \quad t \leq 1 \] Since $t = \sqrt[3]{x}$, \[ \sqrt[3]{x} \leq 1 \quad \Rightarrow \quad x \leq 1 \] Also, $t$ can take any real value, but cube roots are defined for negative $x$. No extra restriction other than $x \leq 1$. Checking lower bound: from cube root properties, minimum $x = -8$ when $t = -2$ is acceptable given no domain restriction. Thus: \[ -8 \leq x \leq 1 \] \[ \boxed{-8 \leq x \leq 1} \]