If 1/a + 1/b + 1/c = 1 and abc = 2, Find ab²c² + a²bc² + a²b²c
If \[ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 \] and \[ abc=2, \] then find \[ ab^2c^2+a^2bc^2+a^2b^2c. \]
Solution
Given:
\[ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 \]
Multiplying both sides by \(abc\):
\[ bc+ca+ab=abc \]
Given:
\[ abc=2 \]
Therefore,
\[ ab+bc+ca=2 \]
Now,
\[ ab^2c^2+a^2bc^2+a^2b^2c \]
Taking \(abc\) common:
\[ =abc(ab+bc+ca) \]
Substituting the values:
\[ =2\times2 \]
\[ =4 \]
Therefore,
\[ \boxed{4} \]