ABC26GN1996 · Surds and Indices

Subject: General Aptitude · Chapter: Surds and Indices · Exam: · Marks: · Difficulty:

If $x^{\frac{1}{3}}+y^{\frac{1}{3}}=z^{\frac{1}{3}}$, then $\left\{(x+y-z)^{3}+27 x y z\right\}$ equals
(a)- 1
(b)0
(c)1
(d)27
Answer
Answer (as printed): B
Explanation
$x^{\frac{1}{3}}+y^{\frac{1}{3}}=z^{\frac{1}{3}} \Rightarrow\left(x^{\frac{1}{3}}+y^{\frac{1}{3}}\right)^{3}=\left(z^{\frac{1}{3}}\right)^{3}$ $$\begin{aligned} & \Rightarrow \quad x+y+3 x^{\frac{1}{3}} y^{\frac{1}{3}}\left(x^{\frac{1}{3}}+y^{\frac{1}{3}}\right)=z \\ & \Rightarrow \quad x+y+3 x^{\frac{1}{3}} y^{\frac{1}{3}} z^{\frac{1}{3}}=z \\ & \Rightarrow \quad x+y-z=-3 x^{\frac{1}{3}} y^{\frac{1}{3}} z^{\frac{1}{3}} \\ & \Rightarrow \quad(x+y-z)^{3}=\left(-3 x^{\frac{1}{3}} y^{\frac{1}{3}} z^{\frac{1}{3}}\right)^{3} \\ & \Rightarrow \quad(x+y-z)^{3}=-27 x y z \Rightarrow(x+y-z)^{3}+27 x y z=0 . \end{aligned}$$

Explanation as extracted from the printed page; notation may be imperfect.

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