Question

If the value of $\dfrac{{{x^2} + 1}}{x} = {3^{\dfrac{1}{3}}}$ Find
(i) $x - \dfrac{1}{x}$
(ii) ${x^3} - \dfrac{1}{{{x^3}}}$

Answer 1

Hint: It is given that the $x + \dfrac{1}{x} = {3^{\dfrac{1}{3}}}$ so for the part (i) $x - \dfrac{1}{x}$ = $y$ , On squaring both we get ${x^2} + \dfrac{1}{{{x^2}}} - 2 = {y^2}$ now find the value of ${x^2} + \dfrac{1}{{{x^2}}}$ from $x + \dfrac{1}{x} = {3^{\dfrac{1}{3}}}$ and put it and get the value of $y$
Now for the part (ii) ${x^3} - \dfrac{1}{{{x^3}}}$ use formula ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$ so from above $a = x$ and $b = \dfrac{1}{x}$ we know the value of ${x^2} + \dfrac{1}{{{x^2}}}$ and $x - \dfrac{1}{x}$ from part (i)

Complete step-by-step answer:
As in the given in the question that is $\dfrac{{{x^2} + 1}}{x} = {3^{\dfrac{1}{3}}}$ , make change in it ,
$x + \dfrac{1}{x} = {3^{\dfrac{1}{3}}}$
So from this we have to find the value of (i) $x - \dfrac{1}{x}$ (ii) ${x^3} - \dfrac{1}{{{x^3}}}$
Now for the part (i) $x - \dfrac{1}{x}$
Let $x - \dfrac{1}{x}$ $ = y$
Now square both side
$\Rightarrow$ ${\left( {x - \dfrac{1}{x}} \right)^2}$ $ = {y^2}$
$\Rightarrow$ ${x^2} + \dfrac{1}{{{x^2}}} - 2 = {y^2}$ .......(i)
Now we have to find the value of ${x^2} + \dfrac{1}{{{x^2}}}$ for finding the value of $y$
So from the question $x + \dfrac{1}{x} = {3^{\dfrac{1}{3}}}$ , on squaring both side ,
$\Rightarrow$ ${x^2} + \dfrac{1}{{{x^2}}} + 2 = {3^{\dfrac{2}{3}}}$
Now put the value of ${x^2} + \dfrac{1}{{{x^2}}} = {3^{\dfrac{2}{3}}} - 2$ in equation (i)
$\Rightarrow$ ${3^{\dfrac{2}{3}}} - 2 - 2 = {y^2}$
$\Rightarrow$ ${3^{\dfrac{2}{3}}} - 4 = {y^2}$
Hence $y = \sqrt {{3^{\dfrac{2}{3}}} - 4} $ i.e.
$\Rightarrow$ $x - \dfrac{1}{x}$ = $y = \sqrt {{3^{\dfrac{2}{3}}} - 4} $
Now for the part (ii) ${x^3} - \dfrac{1}{{{x^3}}}$
we know the formula of ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
So from above $a = x$ and $b = \dfrac{1}{x}$
${x^3} - \dfrac{1}{{{x^3}}}$ = $\left( {x - \dfrac{1}{x}} \right)\left( {{x^2} + x \times \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} \right)$
on solving more
$\Rightarrow$ ${x^3} - \dfrac{1}{{{x^3}}}$ = $\left( {x - \dfrac{1}{x}} \right)\left( {{x^2} + \dfrac{1}{{{x^2}}} + 1} \right)$
So from question we know that $x - \dfrac{1}{x} = \sqrt {{3^{\dfrac{2}{3}}} - 4} $ and above we find that ${x^2} + \dfrac{1}{{{x^2}}} = {3^{\dfrac{2}{3}}} - 2$ , on putting these value we get
$\Rightarrow$ ${x^3} - \dfrac{1}{{{x^3}}}$ = $\left( {\sqrt {{3^{\dfrac{2}{3}}} - 4} } \right)\left( {{3^{\dfrac{2}{3}}} - 2 + 1} \right)$
On multiplying we get ,
$\Rightarrow$ ${x^3} - \dfrac{1}{{{x^3}}}$ = $\left( {\sqrt {{3^{\dfrac{2}{3}}} - 4} } \right)\left( {{3^{\dfrac{2}{3}}} - 1} \right)$
we get , ${x^3} - \dfrac{1}{{{x^3}}}$ = $\left( {\sqrt {{3^{\dfrac{2}{3}}} - 4} } \right)\left( {{3^{\dfrac{2}{3}}} - 1} \right)$
Hence the value of (i) $x - \dfrac{1}{x}$ = $\sqrt {{3^{\dfrac{2}{3}}} - 4} $
(ii) ${x^3} - \dfrac{1}{{{x^3}}}$ = $\left( {\sqrt {{3^{\dfrac{2}{3}}} - 4} } \right)\left( {{3^{\dfrac{2}{3}}} - 1} \right)$

Note: If in the question it is given to find the value of ${x^3} + \dfrac{1}{{{x^3}}}$ then we use the formula that is of ${a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$ so from above $a = x$ and $b = \dfrac{1}{x}$ on putting we get the
${x^3} + \dfrac{1}{{{x^3}}}$ = $\left( {x + \dfrac{1}{x}} \right)\left( {{x^2} - x \times \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} \right)$ so from above $x + \dfrac{1}{x} = {3^{\dfrac{1}{3}}}$ and ${x^2} + \dfrac{1}{{{x^2}}} = {3^{\dfrac{2}{3}}} - 2$ put these value and get the answer .