Subject: General Aptitude · Chapter: Surds and Indices · Exam: · Marks: · Difficulty:
If $a+b+c=0$, then the value of $\left(x^{a}\right)^{a^{2}-b c} \cdot\left(x^{b}\right)^{b^{2}-c a} \cdot\left(x^{c}\right)^{c^{2}-a b}$ is equal to
(a)- 2
(b)- 1
(c)0
(d)1
Answer
Answer (as printed): D
Explanation
$\left(x^{a}\right)^{a^{2}-b c}\left(x^{b}\right)^{b^{2}-c a .}\left(x^{c}\right)^{c^{2}-a b}$ $$\begin{aligned} & =x^{\left[a\left(a^{2}-b c\right)\right] \cdot x^{\left[b\left(b^{2}-c a\right)\right] \cdot x^{\left[c\left(c^{2}-a b\right)\right]}}} \\ & =x^{\left(a^{3}-a b c\right)} \cdot x^{\left(b^{3}-a b c\right)} \cdot x^{\left(c^{3}-a b c\right)} \\ & =x^{\left(a^{3}-a b c+b^{3}-a b c+c^{3}-a b c\right)}=x^{\left(a^{3}+b^{3}+c^{3}-3 a b c\right)} \\ & =x^{(3 a b c-3 a b c)}=x^{0}=1 . \end{aligned} \left[\therefore \quad \text { If } a+b+c=0, a^{3}+b^{3}+c^{3}=3 a b c\right]$$
Explanation as extracted from the printed page; notation may be imperfect.