Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
If $x=a(b-c), y=b(c-a), z=c(a-b)$, then the value of $\left(\frac{x}{a}\right)^{3}+\left(\frac{y}{b}\right)^{3}+\left(\frac{z}{c}\right)^{3}$ is [SSC-CHSL (10 + 2) Exam, 2015]
(a)$\frac{2 x y z}{a b c}$
(b)$\frac{x y z}{a b c}$
(c)0
(d)$\frac{3 x y z}{a b c}$
Answer
Answer (as printed): D
Explanation
Given $x=a(b-c), y=b(c-a) ; z=(a-b)$ $$\begin{aligned} & x=a(b-c) \\ & \Rightarrow \frac{x}{a}=b-c . . \end{aligned}$$ Similarly, $y=b(c-a)$ $$\Rightarrow \frac{y}{b}=c-a \text { (ii) and similarly } z=c(a-b) \frac{z}{c}=c-a$$ Adding (i), (ii) and (iii) we get $$\begin{aligned} & \because \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=b-c+c-a+a-b=0 \\ & \Rightarrow \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0 \\ & \because\left(\frac{x}{a}\right)^{3}+\left(\frac{y}{b}\right)^{3}+\left(\frac{z}{c}\right)^{3} \\ & =3 \times \frac{x}{a} \times \frac{y}{b} \times \frac{z}{c}=\frac{3 x y z}{a b c} \end{aligned}$$ [If $a+b+c=0, a^{3}+b^{3}+c^{3}=3 a b c$ ]
Explanation as extracted from the printed page; notation may be imperfect.