Question

If m,n,p are roots of the equation x^3-3x^2+3x+7=0 and w is complex cube roots of unity find the value of
$\frac{m - 1}{n - 1} + \frac{n - 1}{p - 1} + \frac{p - 1}{m - 1}$


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Answer 1

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

Given Cubic equation is

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

can be rewritten as

$\rm :\longmapsto\: {x}^{3} - {3x}^{2} + 3x - 1 + 8 = 0$

$\rm :\longmapsto\: ({x}^{3} - {3x}^{2} + 3x - 1) + 8 = 0$

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

$\purple{\rm :\longmapsto\: {(x - 1)}^{3} = {( - 2)}^{3}}$

$\rm :\longmapsto\: {\bigg[\dfrac{x - 1}{ - 2} \bigg]}^{3} = 1$

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

We know

$\red{\rm :\longmapsto\:\boxed{\sf{ \: \: \: {\bigg(1\bigg) }^{\dfrac{1}{3} } = 1 \: or \: \omega \: or \: {\omega }^{2} \: \: \: }}}$

So, using this, we get

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

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

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

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

Since, it is given that

$\rm :\longmapsto\:m,n,p \: are \: roots \: of \: {x}^{3} - {3x}^{2} + 3x + 7 = 0$

So,

$\rm :\longmapsto\:m = - 1$

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

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

Now, Consider

$\rm :\longmapsto\:\dfrac{m - 1}{n - 1} + \dfrac{n - 1}{p - 1} + \dfrac{p - 1}{m - 1}$

$\rm \:  =  \: \dfrac{ - 1 - 1}{1 - 2\omega - 1} + \dfrac{1 - 2\omega - 1}{1 - 2 {\omega }^{2} - 1} + \dfrac{1 - {2\omega }^{2} - 1}{ - 1 - 1}$

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

$\rm \:  =  \: \dfrac{1}{\omega} + \dfrac{1}{\omega} + {\omega }^{2}$

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

$\rm \:  =  \: 3{\omega }^{2}$

Hence,

$\rm\implies \:\boxed{\tt{ \dfrac{m - 1}{n - 1} + \dfrac{n - 1}{p - 1} + \dfrac{p - 1}{m - 1} = 3 {\omega }^{2} }}$

Answer 2

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

Given Cubic equation is

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

can be rewritten as

$\rm :\longmapsto\: {x}^{3} - {3x}^{2} + 3x - 1 + 8 = 0$

$\rm :\longmapsto\: ({x}^{3} - {3x}^{2} + 3x - 1) + 8 = 0$

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

$\purple{\rm :\longmapsto\: {(x - 1)}^{3} = {( - 2)}^{3}}$

$\rm :\longmapsto\: {\bigg[\dfrac{x - 1}{ - 2} \bigg]}^{3} = 1$

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

We know

$\red{\rm :\longmapsto\:\boxed{\sf{ \: \: \: {\bigg(1\bigg) }^{\dfrac{1}{3} } = 1 \: or \: \omega \: or \: {\omega }^{2} \: \: \: }}}$

So, using this, we get

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

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

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

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

Since, it is given that

$\rm :\longmapsto\:m,n,p \: are \: roots \: of \: {x}^{3} - {3x}^{2} + 3x + 7 = 0$

So,

$\rm :\longmapsto\:m = - 1$

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

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

Now, Consider

$\rm :\longmapsto\:\dfrac{m - 1}{n - 1} + \dfrac{n - 1}{p - 1} + \dfrac{p - 1}{m - 1}$

$\rm \:  =  \: \dfrac{ - 1 - 1}{1 - 2\omega - 1} + \dfrac{1 - 2\omega - 1}{1 - 2 {\omega }^{2} - 1} + \dfrac{1 - {2\omega }^{2} - 1}{ - 1 - 1}$

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

$\rm \:  =  \: \dfrac{1}{\omega} + \dfrac{1}{\omega} + {\omega }^{2}$

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

$\rm \:  =  \: 3{\omega }^{2}$

Hence,

$\rm\implies \:\boxed{\tt{ \dfrac{m - 1}{n - 1} + \dfrac{n - 1}{p - 1} + \dfrac{p - 1}{m - 1} = 3 {\omega }^{2} }}$