Question

If x^2-4x+1=0, find: (i) x+1/x and (ii) x^3+1/x^3

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: {x}^{2} - 4x + 1 = 0$

can be rewritten as

$\rm :\longmapsto\: {x}^{2} + 1 = 4x$

On dividing both sides by x, we get

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{x} = 4$

$\rm :\longmapsto\:\dfrac{ {x}^{2}}{x} + \dfrac{1}{x} = 4$

$\rm :\longmapsto\:x + \dfrac{1}{x} = 4$

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: x + \dfrac{1}{x} = 4}}}$

On Cubing both sides, we get

$\rm :\longmapsto\: {\bigg[x + \dfrac{1}{x} \bigg]}^{3} = {4}^{3}$

We know,

$\red{\boxed{ \tt{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: }}}$

So, using this identity, we get

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times x \times \dfrac{1}{x}\bigg[x + \dfrac{1}{x} \bigg] = 64$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times 4= 64$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 12= 64$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} }= 64 - 12$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} }= 52$

Hence,

$\red{\rm :\longmapsto\: \boxed{ \tt{ \: {x}^{3} + \dfrac{1}{ {x}^{3} }= 52 \: }}}$

More Identities to know :-

$\blue{\boxed{ \tt{ \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \: }}}$

$\blue{\boxed{ \tt{ \: {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \: }}}$

$\red{\boxed{ \tt{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: }}}$

$\red{\boxed{ \tt{ \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) \: }}}$

$\green{\boxed{ \tt{ \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: }}}$

$\green{\boxed{ \tt{ \:{x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2})\: }}}$

$\green{\boxed{ \tt{ \:{x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2})\: }}}$

$\green{\boxed{ \tt{ \:{x}^{4}- {y}^{4} = (x - y)(x + y)( {x}^{2}+{y}^{2})\: }}}$