Question

Choose the correct answer from the alternatives given.
If a + b + c = 3, $a^2+b^2+c^2=6$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$, where a, b, c are all non-zero, then abc is equal to

• A. 2/3
• B. 3/2
• C. 1/2
• D. 1/3

Answer 1

Answer: (B)

3/2

$(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$
$3^2=6+2(ab+bc+ca)$
$\Rightarrow ab+bc+ca=\frac{96}{2}=\frac{3}{2}$
Now, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
$\Rightarrow \frac{ab+bc+ca}{abc}=1=\Rightarrow \frac{\frac{3}{2}}{abc}=1\Rightarrow abc=\frac{3}{2}$$(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$

$3^2=6+2(ab+bc+ca)$

$\Rightarrow ab+bc+ca=\frac{96}{2}=\frac{3}{2}$

Now, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

$\Rightarrow \frac{ab+bc+ca}{abc}=1=\Rightarrow \frac{\frac{3}{2}}{abc}=1\Rightarrow abc=\frac{3}{2}$