Answer

Verified

419.7k+ views

**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 .

Recently Updated Pages

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Advantages and disadvantages of science

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

10 examples of evaporation in daily life with explanations

Difference Between Plant Cell and Animal Cell

Write a letter to the principal requesting him to grant class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE