Question:
If \[ \frac1a+\frac1b+\frac1c=1 \] and \[ abc=2 \] find:
\[ ab^2c^2+a^2bc^2+a^2b^2c \]
Solution:
Given:
\[ \frac1a+\frac1b+\frac1c=1 \]
Taking LCM:
\[ \frac{bc+ca+ab}{abc}=1 \]
Since \[ abc=2 \]
\[ \frac{ab+bc+ca}{2}=1 \]
\[ ab+bc+ca=2 \]
Now,
\[ ab^2c^2+a^2bc^2+a^2b^2c \]
\[ = abc(ab+bc+ca) \]
Substituting the values:
\[ = 2\times2 \]
\[ =4 \]