Question

If abc = 1, then find the value of:

$\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}$ [4 MARKS]

Answer 1

Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark

Given that,
abc=1
So, $c=1÷ab=(ab{)}^{-1}$
Similarly,
$a=(bc{)}^{-1}$
$b=(ac{)}^{-1}$
And,
$ab=c^{-1}$
$ac=b^{-1}$
$bc=a^{-1}$

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

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

[Multiplying 2nd term by $\frac{b^{-1}}{b^{-1}}$ and 3rd term by $\frac{a}{a}$]

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

$\because \frac{a^m}{a^n}=a^{m-n}$

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

$=1$