Question

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 is non-zero, then $abc$ is

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{3}{2}$
• D. $1$

Answer 1

The correct option is C $\frac{3}{2}$

$a+b+c=3$

$a^2+b^2+c^2=6$

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

Now, $(a+b+c{)}^2=9$

$a^2+b^2+c^2+2(ab+bc+ca)=9$

$6+2(ab+bc+ca)=9$

$(ab+bc+ca)=\frac{3}{2}$

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

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

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