Question

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

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1...\left(1\right) \\ abc=2...\left(2\right) \\ \mathrm{Now}, \\ a{b}^2{c}^2+a^{2}b{c}^2+a^2{b}^{2}c \\ =a^2{b}^2{c}^2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \\ ={\left(abc\right)}^2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \\ ={\left(2\right)}^2\left(1\right)\left(\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\right) \\ =4 \end{gathered}$$

Hence, if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\mathrm{and}abc=2,\mathrm{then}a{b}^2{c}^2+a^{2}b{c}^2+a^2{b}^{2}c=\underline{4}.$