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Question : 50 of 86

Marks: +1, -0

If

a b c = 1

then find the value of

\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}

0
1
$-1$
None.

Solution:

Concept:

Use the condition

abc = 1

to rewrite each term so they combine to give a constant.

Explanation:

Given

abc = 1

, we have

b^{-1} = \frac{1}{b}

,

c^{-1} = \frac{1}{c}

,

a^{-1} = \frac{1}{a}

.

First term:

\frac{1}{1 + a + b^{-1}} = \frac{1}{1 + a + \frac{1}{b}} = \frac{b}{b + ab + 1}

.

Since

ab = \frac{1}{c}

(from

abc=1

), we get

\frac{b}{b + \frac{1}{c} + 1} = \frac{bc}{bc + 1 + c} = \frac{\frac{1}{a}}{\frac{1}{a} + 1 + c} = \frac{1}{1 + a + ac}

.

Second term:

\frac{1}{1 + b + c^{-1}} = \frac{1}{1 + b + \frac{1}{c}} = \frac{c}{c + bc + 1} = \frac{c}{c + \frac{1}{a} + 1} = \frac{ac}{ac + a + 1} = \frac{\frac{1}{b}}{\frac{1}{b} + a + 1} = \frac{1}{1 + ab + b}

? Wait, recalc carefully.

Better: Multiply numerator and denominator by

c

to get

\frac{c}{c + bc + 1}

. Using

bc = \frac{1}{a}

, that becomes

\frac{c}{c + \frac{1}{a} + 1} = \frac{ac}{ac + 1 + a} = \frac{ac}{1 + a + ac}

.

Third term:

\frac{1}{1 + c + a^{-1}} = \frac{1}{1 + c + \frac{1}{a}} = \frac{a}{a + ac + 1} = \frac{a}{1 + a + ac}

.

Now sum all three terms:

\frac{1}{1 + a + ac} + \frac{ac}{1 + a + ac} + \frac{a}{1 + a + ac} = \frac{1 + ac + a}{1 + a + ac} = 1

.

Answer:

1

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