ABC26GN1471 · Square Roots and Cube Roots
Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: 2007 · Marks: · Difficulty:
A rationalising factor of $(\sqrt[3]{9}-\sqrt[3]{3}+1)$ is
(a)$\sqrt[3]{3}-1$
(b)$\sqrt[3]{3}+1$
(c)$\sqrt[3]{9}-1$
(d)$\sqrt[3]{9}+1$
Answer
Explanation
Let $\sqrt[3]{3}=x$. $$\begin{aligned} & \text { Then, }(\sqrt[3]{9}-\sqrt[3]{3}+1)=\left(x^{2}-x+1\right)=\frac{x^{3}+1}{x+1}=\frac{(\sqrt[3]{3})^{3}+1}{(\sqrt[3]{3}+1)} \\ & \Rightarrow(\sqrt[3]{9}-\sqrt[3]{3}+1)(\sqrt[3]{3}+1)=(\sqrt[3]{3})^{3}+1 \\ & \quad=3+1=4 \text {, which is rational. } \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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