Question

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$if \: \alpha \: and \: \beta \: are \: roots \: of \: {x}^{2} + x + 1 = 0$

then form an equation whose roots are

${ \alpha }^{14} + { \beta }^{11} \: and \: { \alpha }^{11} + { \beta }^{14}$

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Answer 1

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

Given that,

$\red{\rm :\longmapsto\: \alpha , \beta \: are \: roots \: of \: {x}^{2} + x + 1 = 0}$

So, using Quadratic formula, we have

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ {1}^{2} - 4(1)(1)} }{2(1)}$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ 1 - 4} }{2}$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ - 3} }{2}$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: i \sqrt{3} }{2}$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: + \: i \sqrt{3} }{2} \: \: or \: \: \dfrac{ - 1 \: - \: i \sqrt{3} }{2}$

$\bf\implies \:x \: = \: \omega \: \: or \: \: {\omega }^{2}$

So, it implies,

$\rm \implies\: \alpha = \omega$

and

$\rm \implies\: \beta = {\omega }^{2}$

Now, Consider

$\red{\rm :\longmapsto\: { \alpha }^{14} + { \beta }^{11}}$

$\rm \:  =  \: {\omega }^{14} + {( {\omega }^{2} )}^{11}$

$\rm \:  =  \: {\omega }^{12 + 2} + {\omega }^{22}$

$\rm \:  =  \: {\omega }^{12} \times {\omega }^{2} + {\omega }^{21} \times \omega$

$\rm \:  =  \: {\omega }^{2} + \omega$

$\rm \:  =  \: - 1$

Now, Consider

$\red{\rm :\longmapsto\: { \alpha }^{11} + { \beta }^{14}}$

$\rm \:  =  \: {\omega }^{11} + {( {\omega }^{2}) }^{14}$

$\rm \:  =  \: {\omega }^{9 + 2} + {\omega }^{28}$

$\rm \:  =  \: {\omega }^{9} \times {\omega }^{2} + {\omega }^{27} \times \omega$

$\rm \:  =  \: {\omega }^{2} + \omega$

$\rm \:  =  \: - 1$

Now, Consider

Sum of roots

$\red{\rm :\longmapsto\: { \alpha }^{14} + { \beta }^{11} + { \alpha }^{11} + { \beta }^{14} }$

$\rm \:  =  \: ( - 1) + ( - 1)$

$\rm \:  =  \: - 2$

Product of roots

$\red{\rm :\longmapsto\: { (\alpha }^{14} + { \beta }^{11}) \times ({ \alpha }^{11} + { \beta }^{14} )}$

$\rm \:  =  \: ( - 1) \times ( - 1)$

$\rm \:  =  \: 1$

So, Required Quadratic equation is

$\red{\rm :\longmapsto\: {x}^{2} - (sum \: of \: roots)x + (product \: of \: roots) = 0}$

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

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

$\rm :\longmapsto\: {(x + 1)}^{2} = 0$

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Formula Used :-

$\boxed{\tt{ {\omega }^{3} = 1 \: }}$

$\boxed{\tt{ 1 + \omega + {\omega }^{2} = 0 \: }}$