Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
If $a+b+c=6$ and $a b+b c+c a=10$ then the value of $a^{3}+b^{3}+c^{3}-3 a b c$ is [SSC-CHSL (10 + 2) Exam, 2015]
(a)36
(b)48
(c)42
(d)40
Answer
Answer (as printed): A
Explanation
Given $$\begin{aligned} & a+b+c=6 \\ & a b+b c+c a=10 \end{aligned} \begin{aligned} & \therefore(a+b+c)^{2}=36 \\ & \Rightarrow a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a=36 \\ & \Rightarrow a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=36 \\ & \Rightarrow a^{2}+b^{2}+c^{2}+2 \times 10=36 \\ & \Rightarrow a^{2}+b^{2}+c^{2}=16 \end{aligned}$$ As we know $\frac{a^{3}+b^{3}+c^{3}-3 a b c}{a^{2}+b^{2}+c^{2}-a b-b c-c a}=(a+b+c)$ $$\begin{aligned} & \frac{a^{3}+b^{3}+c^{3}-3 a b c}{16-(a b+b c+c a)}=6 \\ & \Rightarrow \frac{a^{3}+b^{3}+c^{3}-3 a b c}{16-10}=6 \\ & \Rightarrow a^{3}+b^{3}+c^{3}-3 a b c=6 \times 6 \\ & \Rightarrow a^{3}+b^{3}+c^{3}-3 a b c=36 \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.