Question

X=1/4-x find the value of
1.x+1/X
2.x3+1/x3
3.x6+1/x6​

Answer 1

$\mathfrak{\underline{\underline{Answer}}}$

(i)$\mathbf{ x + \frac{1}{x}= 4}$

(ii) $\mathbf{ {x}^{3}+ \frac{1}{{x}^{3}}= 52}$

(iii)$\mathbf{ {x}^{6}+ \frac{1}{x^{6}}= 2702}$

$\mathfrak{\underline{\underline{Explanation:-}}}$

Given,

x = $\dfrac{1}{4-x}$

______________________________

(i) x + $\dfrac{1}{x}$

$\mathbf{x = \frac{1}{4-x} }$

$\mathbf{ \frac{1}{x}= 4-x}$

$\mathbf{ x + \frac{1}{x}= 4}$

_______________________________

(ii) x³ + $\frac{1}{{x}^{3}}$

As we got,

$\mathbf{ x +\frac{1}{x}= 4}$

Now, by cubing on both sides

$\mathbf{ (x + \frac{1}{{x}})^{3}= 4^{3}}$

$\boxed{{(a+b)}^{3}= {a}^{3}+{b}^{3}+3ab(a+b)}$

$\mathbf{ {x}^{3}+ \frac{1}{{x}^{3}}+3x ×\frac{1}{x}(x+\frac{1}{x})= 64}$

$\mathbf{ {{x}^{3}}+ \frac{1}{{x}^{3}}+3(4)= 64}$

$\mathbf{ {x}^{3}+ \frac{1}{{x}^{3}}+12= 64}$

$\mathbf{ {x}^{3}+ \frac{1}{{x}^{3}}= 64-12}$

$\mathbf{ {x}{3}+ \frac{1}{{x}^{3}}= 52}$

_______________________________

(iii) ${x}^{6}+\frac{1}{{x}^{6}}$

As we have,

$\mathbf{ {x}^{3}+ \frac{1}{{x}^{3}}= 52}$

Now, by squaring on both sides

$\mathbf{ ({x}^{3}+ \frac{1}{{x}^{3}})^{2}= {52}^{2}}$

$\boxed{{(a+b)}^{2} = {a}^{2}+{b}^{2}+2ab}$

$\mathbf{({{x}^{3}})^{2}+ \frac{1}{({x}^{3})^{2}}+2( {x}^{3} \times{\frac{1}{{x}^{3}}})= 2704}$

$\mathbf{ {x}^{6}+ \frac{1}{{x}^{6}}+2= 2704}$

$\mathbf{ {x}^{6}+ \frac{1}{{x}^{6}}= 2704-2}$

$\mathbf{ {x}^{6}+ \frac{1}{{x}^{6}}= 2702}$

Answer 2

$\mathfrak{\large{\underline{\underline{Answer :-}}}}$

$\boxed{\sf{(i)\:x + \dfrac{1}{x} = 4}}$

$\boxed{\sf{(ii)\:x^3 + \dfrac{1}{x^3} = 52}}$

$\boxed{\sf{(iii)\:{x}^{6} + \dfrac{1}{ {x}^{6} } = 2702}}$

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

$\bf{x = \dfrac{1}{4 - x} }$

$\bf{\implies{ \dfrac{1}{x} = 4 - x }}$

$\bf{\implies{\dfrac{1}{x} + x= 4}}$

$\bf{\implies{x + \dfrac{1}{x} = 4}}$

$\boxed{\sf{(i)\:x + \dfrac{1}{x} = 4}}$

Now on cubing on both sides

$\bf{\implies{{\left( \dfrac{1}{x} + x \right)}^{3} = {4}^{3} }}$

We know, that (a + b)³ = a³ + b³ + 3ab(a + b)

Here a = x , b = 1/x

By substituting the values in the above identity we have,

$\bf{\implies{{x}^{3} + \dfrac{1}{ {x}^{3}} + 3 \times x \times \dfrac{1}{x} \left(x + \dfrac{1}{x} \right) = 64}}$

$\bf{\implies{{x}^{3} + \dfrac{1}{ {x}^{3}} + 3 \left( x + \dfrac{1}{x} \right) = 64}}$

We know that x + 1/x = 4 so substitute the value in the equation.

$\bf{\implies{{x}^{3} + \dfrac{1}{ {x}^{3}} + 3(4) = 64}}$

$\bf{\implies{{x}^{3} + \dfrac{1}{ {x}^{3}} + 12 = 64}}$

$\bf{\implies{{x}^{3} + \dfrac{1}{ {x}^{3}} = 64 - 12}}$

$\boxed{\sf{(ii)\:x^3 + \dfrac{1}{x^3} = 52}}$

Now squaring on both sides

$\bf{{ \left( {x}^{3} + \dfrac{1}{ {x}^{3}} \right)}^{2} = {52}^{2} }$

We know that, (a + b)² = a² + b² + 2ab

Here a = x³ , b = 1/x³

By substituting the values in the above identity we have,

$\bf{{({x}^{3})}^{2} + {({(\dfrac{1}{x})}^{3})}^{2} + 2 \times {x}^{3} \times \dfrac{1}{ {x}^{3} } = 2704}$

$\bf{\implies{{x}^{3 \times 2} + {( \dfrac{1}{x})}^{3 \times 2} + 2 = 2704}}$

$\bf{\implies{{x}^{6} + {( \dfrac{1}{x})}^{6}= 2704 - 2}}$

$\bf{\implies{{x}^{6} + \dfrac{1}{ {x}^{6} } = 2702}}$

$\boxed{\sf{(iii)\:{x}^{6} + \dfrac{1}{ {x}^{6} } = 2702}}$

$\mathfrak{\large{\underline{\underline{Identities\:Used:-}}}}$

[1] (a + b)³ = a³ + b³ + 3ab(a + b)

[2] (a + b)³ = a² + b² + 2ab