# Solve the following equations:

$\dfrac{{{x^3} + 1}}{{{x^2} - 1}} = x + \sqrt {\dfrac{6}{x}} .$

Last updated date: 27th Mar 2023

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Answer

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Hint: - Use \[{{\text{a}}^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)\], \[\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)\]

As we know \[{{\text{a}}^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)\], and \[\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)\]

So, apply these properties

\[

\Rightarrow \dfrac{{{x^3} + 1}}{{{x^2} - 1}} = x + \sqrt {\dfrac{6}{x}} . \\

\Rightarrow \dfrac{{\left( {x + 1} \right)\left( {{x^2} + 1 - x} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = x + \sqrt {\dfrac{6}{x}} \\

\]

\[ \Rightarrow \left( {x + 1} \right)\]Is canceled out from the numerator and denominator.

\[

\Rightarrow \dfrac{{\left( {{x^2} + 1 - x} \right)}}{{\left( {x - 1} \right)}} = x + \sqrt {\dfrac{6}{x}} \\

\Rightarrow \dfrac{{\left( {{x^2} + 1 - x} \right)}}{{\left( {x - 1} \right)}} - x = \sqrt {\dfrac{6}{x}} \Rightarrow \dfrac{{\left( {{x^2} + 1 - x} \right) - {x^2} + x}}{{x - 1}} = \sqrt {\dfrac{6}{x}} \Rightarrow \dfrac{1}{{x - 1}} = \sqrt {\dfrac{6}{x}} \\

\]

By, cross multiplication

\[ \Rightarrow 1 \times \sqrt x = \left( {x - 1} \right)\sqrt 6 \]

Now, squaring on both sides

\[

\Rightarrow {\left( {\sqrt x } \right)^2} = {\left( {x - 1} \right)^2}{\left( {\sqrt 6 } \right)^2} \Rightarrow x = 6\left( {{x^2} + 1 - 2x} \right) \\

\Rightarrow 6{x^2} - 13x + 6 = 0 \\

\Rightarrow 6{x^2} - 9x - 4x + 6 = 0 \\

\Rightarrow 3x\left( {2x - 3} \right) - 2\left( {2x - 3} \right) = 0 \\

\Rightarrow \left( {2x - 3} \right)\left( {3x - 2} \right) = 0 \\

\Rightarrow \left( {2x - 3} \right) = 0{\text{, and }}\left( {3x - 2} \right) = 0 \\

\Rightarrow x = \dfrac{3}{2},{\text{ and }}x = \dfrac{2}{3} \\

\]

So, this is the required answer.

Note: -In such types of questions always remember the basic formulas which are stated above then after cross multiplication factorize the equation and simplify the equation we will get the required solution of the equation.

As we know \[{{\text{a}}^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)\], and \[\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)\]

So, apply these properties

\[

\Rightarrow \dfrac{{{x^3} + 1}}{{{x^2} - 1}} = x + \sqrt {\dfrac{6}{x}} . \\

\Rightarrow \dfrac{{\left( {x + 1} \right)\left( {{x^2} + 1 - x} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = x + \sqrt {\dfrac{6}{x}} \\

\]

\[ \Rightarrow \left( {x + 1} \right)\]Is canceled out from the numerator and denominator.

\[

\Rightarrow \dfrac{{\left( {{x^2} + 1 - x} \right)}}{{\left( {x - 1} \right)}} = x + \sqrt {\dfrac{6}{x}} \\

\Rightarrow \dfrac{{\left( {{x^2} + 1 - x} \right)}}{{\left( {x - 1} \right)}} - x = \sqrt {\dfrac{6}{x}} \Rightarrow \dfrac{{\left( {{x^2} + 1 - x} \right) - {x^2} + x}}{{x - 1}} = \sqrt {\dfrac{6}{x}} \Rightarrow \dfrac{1}{{x - 1}} = \sqrt {\dfrac{6}{x}} \\

\]

By, cross multiplication

\[ \Rightarrow 1 \times \sqrt x = \left( {x - 1} \right)\sqrt 6 \]

Now, squaring on both sides

\[

\Rightarrow {\left( {\sqrt x } \right)^2} = {\left( {x - 1} \right)^2}{\left( {\sqrt 6 } \right)^2} \Rightarrow x = 6\left( {{x^2} + 1 - 2x} \right) \\

\Rightarrow 6{x^2} - 13x + 6 = 0 \\

\Rightarrow 6{x^2} - 9x - 4x + 6 = 0 \\

\Rightarrow 3x\left( {2x - 3} \right) - 2\left( {2x - 3} \right) = 0 \\

\Rightarrow \left( {2x - 3} \right)\left( {3x - 2} \right) = 0 \\

\Rightarrow \left( {2x - 3} \right) = 0{\text{, and }}\left( {3x - 2} \right) = 0 \\

\Rightarrow x = \dfrac{3}{2},{\text{ and }}x = \dfrac{2}{3} \\

\]

So, this is the required answer.

Note: -In such types of questions always remember the basic formulas which are stated above then after cross multiplication factorize the equation and simplify the equation we will get the required solution of the equation.

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