Question

$if \: \alpha \: \beta \: \gamma \: are \: roots \: of \: {x}^{3} - {3x}^{2} + 3x + 7 = 0$
and

omega is cube roots of unity, then

$( \alpha - 1)( \beta - 1) + ( \beta - 1)( \gamma - 1)( \gamma - 1)( \alpha - 1) =$

EVALUATE the above

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Answer 1

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

Given that,

$\rm :\longmapsto\:1,\omega , {\omega }^{2} \: are \: cube \: roots \: of \: unity.$

So,

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

and

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

Also, given that,

$\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: the \: roots \: of \: {x}^{3} - {3x}^{2} + 3x + 7 = 0$

Consider,

$\rm :\longmapsto\: {x}^{3} - {3x}^{2} + 3x + 7 = 0$

$\rm :\longmapsto\: {x}^{3} - {3x}^{2} + 3x = - 7$

$\rm :\longmapsto\: {x}^{3} - {3x}^{2} + 3x - 1 = - 7 - 1$

$\rm \implies\: {(x - 1)}^{3} = - 8$

$\rm \implies\:x - 1 = {\bigg( - 8\bigg) }^{\dfrac{1}{3} }$

$\rm \implies\:x - 1 = - 2{\bigg(1\bigg) }^{\dfrac{1}{3} }$

$\rm \implies\:x - 1 = - 2, \: - 2\omega , \: - {2\omega }^{2}$

As, it is given that

$\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: the \: roots \: of \: {x}^{3} - {3x}^{2} + 3x + 7 = 0$

So, Let assume that

$\rm :\longmapsto\: \alpha - 1 = - 2$

$\rm :\longmapsto\: \beta - 1 = - 2\omega$

$\rm :\longmapsto\: \gamma - 1 = - {2\omega }^{2}$

Now, Consider

$\rm :\longmapsto\:( \alpha - 1)(\beta - 1) + ( \beta - 1)( \gamma - 1) + (\gamma - 1)( \alpha - 1)$

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

$\rm \:  =  \: {4\omega }+ {4\omega }^{3} + {4\omega }^{2}$

$\rm \:  =  \: 4\omega (1 + {\omega }^{2} + \omega )$

$\rm \:  =  \: 4\omega \times 0$

$\rm \:  =  \: 0$

Hence,

$\boxed{\tt{ ( \alpha - 1)(\beta - 1) + ( \beta - 1)( \gamma - 1) + (\gamma - 1)( \alpha - 1) = 0}}$

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Cube Roots of unity

$\rm :\longmapsto\:x = {\bigg(1\bigg) }^{\dfrac{1}{3} }$

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

$\rm :\longmapsto\: {x}^{3} - 1 = 0$

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

$\bf\implies \:x = 1$

Or

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

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

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

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

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

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

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

$\bf\implies \:x = 1 \: \: or \: \: \omega \: \: or \: \: {\omega }^{2}$

where,

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

and

$\rm :\longmapsto\: {\omega }^{2} = \dfrac{ - 1 \: - \: i\sqrt{3} }{2}$