If $a + b + c = abc$ , show that $\dfrac{{a\left( {{b^2}{c^2} - 1} \right)}}{{bc + 1}} + \dfrac{{b\left( {{c^2}{a^2} - 1} \right)}}{{ca + 1}} + \dfrac{{c\left( {{a^2}{b^2} - 1} \right)}}{{ab + 1}} = 2abc$.
Answer
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Hint: Simplify the expression using algebraic identity ${x^2} - {y^2} = \left( {x - y} \right)\left( {x + y} \right)$.
Given, ${\text{ }}a + b + c = abc{\text{ }} \to {\text{(1)}}$
To prove- $\dfrac{{a\left( {{b^2}{c^2} - 1} \right)}}{{bc + 1}} + \dfrac{{b\left( {{c^2}{a^2} - 1} \right)}}{{ca + 1}} + \dfrac{{c\left( {{a^2}{b^2} - 1} \right)}}{{ab + 1}} = 2abc$
Now, let us take the LHS of the equation that we need to prove and simply it.
i.e., \[{\text{LHS}} = \dfrac{{a\left( {{b^2}{c^2} - 1} \right)}}{{bc + 1}} + \dfrac{{b\left( {{c^2}{a^2} - 1} \right)}}{{ca + 1}} + \dfrac{{c\left( {{a^2}{b^2} - 1} \right)}}{{ab + 1}} = \dfrac{{a\left[ {{{\left( {bc} \right)}^2} - {1^2}} \right]}}{{bc + 1}} + \dfrac{{b\left[ {{{\left( {ca} \right)}^2} - {1^2}} \right]}}{{ca + 1}} + \dfrac{{c\left[ {{{\left( {ab} \right)}^2} - {1^2}} \right]}}{{ab + 1}}\]
Using the formula, ${x^2} - {y^2} = \left( {x - y} \right)\left( {x + y} \right)$
$
\Rightarrow {\text{LHS}} = \dfrac{{a\left( {bc - 1} \right)\left( {bc + 1} \right)}}{{bc + 1}} + \dfrac{{b\left( {ca - 1} \right)\left( {ca + 1} \right)}}{{ca + 1}} + \dfrac{{c\left( {ab - 1} \right)\left( {ab + 1} \right)}}{{ab + 1}} \\
\Rightarrow {\text{LHS}} = \dfrac{{a\left( {bc - 1} \right)}}{1} + \dfrac{{b\left( {ca - 1} \right)}}{1} + \dfrac{{c\left( {ab - 1} \right)}}{1} \\
\Rightarrow {\text{LHS}} = abc - a + abc - b + abc - c = 3abc - \left( {a + b + c} \right) \\
$
Using equation (1), we get
$ \Rightarrow {\text{LHS}} = 3abc - abc = 2abc = {\text{RHS}}$
Clearly, the above equation shows that the LHS taken is equal to the RHS of the equation which needs to be proved.
Hence, $\dfrac{{a\left( {{b^2}{c^2} - 1} \right)}}{{bc + 1}} + \dfrac{{b\left( {{c^2}{a^2} - 1} \right)}}{{ca + 1}} + \dfrac{{c\left( {{a^2}{b^2} - 1} \right)}}{{ab + 1}} = 2abc$
Note- In these types of problems, the LHS of the equation which needs to be proved is simplified using the given data in such a way that it ends up becoming equal to the RHS of the equation.
Given, ${\text{ }}a + b + c = abc{\text{ }} \to {\text{(1)}}$
To prove- $\dfrac{{a\left( {{b^2}{c^2} - 1} \right)}}{{bc + 1}} + \dfrac{{b\left( {{c^2}{a^2} - 1} \right)}}{{ca + 1}} + \dfrac{{c\left( {{a^2}{b^2} - 1} \right)}}{{ab + 1}} = 2abc$
Now, let us take the LHS of the equation that we need to prove and simply it.
i.e., \[{\text{LHS}} = \dfrac{{a\left( {{b^2}{c^2} - 1} \right)}}{{bc + 1}} + \dfrac{{b\left( {{c^2}{a^2} - 1} \right)}}{{ca + 1}} + \dfrac{{c\left( {{a^2}{b^2} - 1} \right)}}{{ab + 1}} = \dfrac{{a\left[ {{{\left( {bc} \right)}^2} - {1^2}} \right]}}{{bc + 1}} + \dfrac{{b\left[ {{{\left( {ca} \right)}^2} - {1^2}} \right]}}{{ca + 1}} + \dfrac{{c\left[ {{{\left( {ab} \right)}^2} - {1^2}} \right]}}{{ab + 1}}\]
Using the formula, ${x^2} - {y^2} = \left( {x - y} \right)\left( {x + y} \right)$
$
\Rightarrow {\text{LHS}} = \dfrac{{a\left( {bc - 1} \right)\left( {bc + 1} \right)}}{{bc + 1}} + \dfrac{{b\left( {ca - 1} \right)\left( {ca + 1} \right)}}{{ca + 1}} + \dfrac{{c\left( {ab - 1} \right)\left( {ab + 1} \right)}}{{ab + 1}} \\
\Rightarrow {\text{LHS}} = \dfrac{{a\left( {bc - 1} \right)}}{1} + \dfrac{{b\left( {ca - 1} \right)}}{1} + \dfrac{{c\left( {ab - 1} \right)}}{1} \\
\Rightarrow {\text{LHS}} = abc - a + abc - b + abc - c = 3abc - \left( {a + b + c} \right) \\
$
Using equation (1), we get
$ \Rightarrow {\text{LHS}} = 3abc - abc = 2abc = {\text{RHS}}$
Clearly, the above equation shows that the LHS taken is equal to the RHS of the equation which needs to be proved.
Hence, $\dfrac{{a\left( {{b^2}{c^2} - 1} \right)}}{{bc + 1}} + \dfrac{{b\left( {{c^2}{a^2} - 1} \right)}}{{ca + 1}} + \dfrac{{c\left( {{a^2}{b^2} - 1} \right)}}{{ab + 1}} = 2abc$
Note- In these types of problems, the LHS of the equation which needs to be proved is simplified using the given data in such a way that it ends up becoming equal to the RHS of the equation.
Last updated date: 18th Sep 2023
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