Question

If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ and $abc=2,$ then $a{b}^2{c}^2+a^{2}b{c}^2+a^2{b}^{2}c$ is

Answer 1

We have,

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ and $abc=2$

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

$bc+ca+ab=abc$

On multiply by $abc$ both sides, we get

$abc(bc+ca+ab)=abc\times abc$

$a{b}^2{c}^2+a^{2}b{c}^2+a^2{b}^{2}c=(abc{)}^2$

$a{b}^2{c}^2+a^{2}b{c}^2+a^2{b}^{2}c=2^2=4$

Hence, this is the answer.