If \(\mathrm{a}+\mathrm{b}+\mathrm{c}=3, \mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=6\) and \(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}+\frac{1}{\mathrm{c}}=1\), where \(\mathrm{a}\), \(\mathrm{b}, \mathrm{c}\) are all non-zero, then ‘abc’ is equal to
(a) \(\frac{2}{3}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{3}\)
If a+b+c=3, a^2+b^2+c^2=6, and 1/a+1/b+1/c=1, where a, b, c are all non-zero, then ‘abc’ is equal to (a) 2/3 (b) 3/2 (c) 1/2 (d) 1/3
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Given the system of equations:
1. \(a + b + c = 3\)
2. \(a^2 + b^2 + c^2 = 6\)
3. \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\)
We want to find the value of \(abc\).
To solve this, let’s first analyze the given information. The third equation can be written as:
\[
\frac{ab + ac + bc}{abc} = 1
\]
From this, we get:
\[
ab + ac + bc = abc
\]
From the first equation, we know the sum of \(a\), \(b\), and \(c\). To find \(ab + ac + bc\), we can square the first equation and compare it to the second equation. Squaring the first equation gives us:
\[
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) = 3^2 = 9
\]
Given \(a^2 + b^2 + c^2 = 6\) from the second equation, we can substitute that into our squared first equation to find \(ab + ac + bc\):
\[
6 + 2(ab + ac + bc) = 9
\]
Solving for \(ab + ac + bc\):
\[
2(ab + ac + bc) = 3
\]
\[
ab + ac + bc = \frac{3}{2}
\]
Given \(ab + ac + bc = abc\), we have:
\[
abc = \frac{3}{2}
\]
Therefore, the value of \(abc\) is \(\frac{3}{2}\).
Answer: (b) \(\frac{3}{2}\).