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Hint- Replace the value of x as its inverse in the given equation and proceed with sum and product of roots.

Given equation is \[{x^4} + \dfrac{8}{9}{x^2} + 1 = 3{x^3} + 3x \Rightarrow {x^4} + 3{x^3} + \dfrac{{8{x^2}}}{9} + 3x + 1 = 0{\text{ }} \to {\text{(1)}}\]

Let us replace \[x\] by \[\dfrac{1}{x}\] in the given equation, we get

\[{\left( {\dfrac{1}{x}} \right)^4} + \dfrac{8}{9}{\left( {\dfrac{1}{x}} \right)^2} + 1 = 3{\left( {\dfrac{1}{x}} \right)^3} + \dfrac{3}{x} \Rightarrow \dfrac{1}{{{x^4}}} + \dfrac{8}{{9{x^2}}} + 1 = \dfrac{3}{{{x^3}}} + \dfrac{3}{x}\]

Taking \[{\text{9}}{x^4}\] as the LCM on the LHS and \[{x^3}\] as the LCM on the RHS of the above equation

\[

\Rightarrow \dfrac{{9 + 8{x^2} + 9{x^4}}}{{9{x^4}}} = \dfrac{{3 + 3{x^2}}}{{{x^3}}} \Rightarrow \dfrac{{9 + 8{x^2} + 9{x^4}}}{{9x}} = 3 + 3{x^2} \Rightarrow \dfrac{{9 + 8{x^2} + 9{x^4}}}{9} = x\left( {3 + 3{x^2}} \right) \\

\Rightarrow 1 + \dfrac{{8{x^2}}}{9} + {x^4} = 3x + 3{x^3} \Rightarrow {x^4} + 3{x^3} + \dfrac{{8{x^2}}}{9} + 3x + 1 = 0 \\

\]

Clearly, the above equation which is obtained by replacing \[x\] by \[\dfrac{1}{x}\] in the given equation is the same as the given equation.

As, we know that in case of four degree polynomial (having two roots as \[\alpha \], \[\beta \]) if \[x\] is replaced by \[\dfrac{1}{x}\] and the fourth degree polynomial comes out to be same as the previous one then the other two roots of that polynomial will be \[\dfrac{1}{\alpha }\] and \[\dfrac{1}{\beta }\].

Also, for any general fourth degree polynomial \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\]

Sum of all the roots\[ = - \dfrac{b}{a}\]

Sum of product of different roots taken two at a time\[ = \dfrac{c}{a}\]

According to the given equation (1), we can say \[a = 1,{\text{ }}b = 3,{\text{ }}c = \dfrac{8}{9},{\text{ }}d = 3\] and \[e = 1\].

Therefore, Sum of all the roots of the given equation (1) is given by \[\alpha + \beta + \dfrac{1}{\alpha } + \dfrac{1}{\beta } = - \dfrac{3}{1} = - 3{\text{ }} \to {\text{(2)}}\]

Also, sum of product of different roots taken two at a time of the given equation (1) is given by

\[\alpha \beta {\text{ + }}\alpha {\text{.}}\dfrac{1}{\alpha }{\text{ + }}\alpha {\text{.}}\dfrac{1}{\beta }{\text{ + }}\beta {\text{.}}\dfrac{1}{\alpha }{\text{ + }}\beta {\text{.}}\dfrac{1}{\beta }{\text{ + }}\dfrac{1}{\alpha }{\text{.}}\dfrac{1}{\beta }{\text{ }} = \dfrac{{\dfrac{8}{9}}}{1} = \dfrac{8}{9}\]

\[

\Rightarrow \alpha \beta {\text{ + }}1{\text{ + }}\alpha {\text{.}}\dfrac{1}{\beta }{\text{ + }}\beta {\text{.}}\dfrac{1}{\alpha }{\text{ + }}1{\text{ + }}\dfrac{1}{\alpha }{\text{.}}\dfrac{1}{\beta }{\text{ }} = \dfrac{8}{9} \Rightarrow \alpha \beta {\text{ + }}\alpha {\text{.}}\dfrac{1}{\beta }{\text{ + }}\beta {\text{.}}\dfrac{1}{\alpha }{\text{ + }}\dfrac{1}{\alpha }{\text{.}}\dfrac{1}{\beta } = \dfrac{8}{9} - 2 = - \dfrac{{10}}{9} \\

\Rightarrow \alpha \left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right){\text{ + }}\dfrac{1}{\alpha }\left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right) = - \dfrac{{10}}{9} \Rightarrow \left( {\alpha + \dfrac{1}{\alpha }} \right)\left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right) = - \dfrac{{10}}{9}{\text{ }} \to {\text{(3)}} \\

\]

Equation (2) can be rearranged as \[\left( {\alpha + \dfrac{1}{\alpha }} \right) = - 3 - \left( {\beta + \dfrac{1}{\beta }} \right)\]

Put the value of \[\left( {\alpha + \dfrac{1}{\alpha }} \right)\] in equation (3), we get

\[\left[ { - 3 - \left( {\beta + \dfrac{1}{\beta }} \right)} \right]\left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right) = - \dfrac{{10}}{9}\]

Let \[\left( {\beta + \dfrac{1}{\beta }} \right) = t\]

\[

\Rightarrow \left[ { - 3 - t} \right]t = - \dfrac{{10}}{9} \Rightarrow - \left( {3 + t} \right)t = - \dfrac{{10}}{9} \Rightarrow \left( {3 + t} \right)t = \dfrac{{10}}{9} \Rightarrow 9{t^2} + 27t - 10 = 0 \\

\Rightarrow 9{t^2} + 27t - 10 = 0 \Rightarrow 9{t^2} - 3t + 30t - 10 = 0 \Rightarrow 3t\left( {3t - 1} \right) + 10\left( {3t - 1} \right) = 0 \\

\Rightarrow \left( {3t - 1} \right)\left( {3t + 10} \right) = 0 \\

\]

i.e., Either \[3t - 1 = 0\] or \[3t + 10 = 0\]

\[

\Rightarrow t = \dfrac{1}{3} \Rightarrow \beta + \dfrac{1}{\beta } = \dfrac{1}{3} \Rightarrow \dfrac{{{\beta ^2} + 1}}{\beta } = \dfrac{1}{3} \Rightarrow 3{\beta ^2} + 3 = \beta \Rightarrow 3{\beta ^2} - \beta + 3 = 0 \Rightarrow \beta = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \times 3 \times 3} }}{{2 \times 3}} \\

\Rightarrow \beta = \dfrac{{1 \pm \sqrt {1 - 36} }}{6} \Rightarrow \beta = \dfrac{{1 \pm \sqrt { - 35} }}{6} \Rightarrow \beta = \dfrac{{1 \pm i\sqrt {35} }}{6} \\

\]

or \[

t = - \dfrac{{10}}{3} \Rightarrow \beta + \dfrac{1}{\beta } = - \dfrac{{10}}{3} \Rightarrow \dfrac{{{\beta ^2} + 1}}{\beta } = - \dfrac{{10}}{3} \Rightarrow 3{\beta ^2} + 3 = - 10\beta \Rightarrow 3{\beta ^2} + 10\beta + 3 = 0 \Rightarrow 3{\beta ^2} + 9\beta + \beta + 3 = 0 \\

\Rightarrow 3\beta \left( {\beta + 3} \right) + 1\left( {\beta + 3} \right) = 0 \Rightarrow \left( {3\beta + 1} \right)\left( {\beta + 3} \right) = 0 \Rightarrow \beta = - \dfrac{1}{3}, - 3 \\

\]

Using equation (2) put the value of \[\beta \], we will get the value of \[\alpha \]

\[ \Rightarrow \alpha = - \dfrac{1}{3}, - 3\] or \[\alpha = \dfrac{{1 \pm i\sqrt {35} }}{6}\]

Therefore, all the roots of the given equation are \[ - \dfrac{1}{3}, - 3,\dfrac{{1 \pm i\sqrt {35} }}{6}\].

Note- These types of problems can be solved by somehow checking for some properties regarding roots of a polynomial and then finding out an appropriate relation between the roots and hence solving further to get the values of these roots.

Given equation is \[{x^4} + \dfrac{8}{9}{x^2} + 1 = 3{x^3} + 3x \Rightarrow {x^4} + 3{x^3} + \dfrac{{8{x^2}}}{9} + 3x + 1 = 0{\text{ }} \to {\text{(1)}}\]

Let us replace \[x\] by \[\dfrac{1}{x}\] in the given equation, we get

\[{\left( {\dfrac{1}{x}} \right)^4} + \dfrac{8}{9}{\left( {\dfrac{1}{x}} \right)^2} + 1 = 3{\left( {\dfrac{1}{x}} \right)^3} + \dfrac{3}{x} \Rightarrow \dfrac{1}{{{x^4}}} + \dfrac{8}{{9{x^2}}} + 1 = \dfrac{3}{{{x^3}}} + \dfrac{3}{x}\]

Taking \[{\text{9}}{x^4}\] as the LCM on the LHS and \[{x^3}\] as the LCM on the RHS of the above equation

\[

\Rightarrow \dfrac{{9 + 8{x^2} + 9{x^4}}}{{9{x^4}}} = \dfrac{{3 + 3{x^2}}}{{{x^3}}} \Rightarrow \dfrac{{9 + 8{x^2} + 9{x^4}}}{{9x}} = 3 + 3{x^2} \Rightarrow \dfrac{{9 + 8{x^2} + 9{x^4}}}{9} = x\left( {3 + 3{x^2}} \right) \\

\Rightarrow 1 + \dfrac{{8{x^2}}}{9} + {x^4} = 3x + 3{x^3} \Rightarrow {x^4} + 3{x^3} + \dfrac{{8{x^2}}}{9} + 3x + 1 = 0 \\

\]

Clearly, the above equation which is obtained by replacing \[x\] by \[\dfrac{1}{x}\] in the given equation is the same as the given equation.

As, we know that in case of four degree polynomial (having two roots as \[\alpha \], \[\beta \]) if \[x\] is replaced by \[\dfrac{1}{x}\] and the fourth degree polynomial comes out to be same as the previous one then the other two roots of that polynomial will be \[\dfrac{1}{\alpha }\] and \[\dfrac{1}{\beta }\].

Also, for any general fourth degree polynomial \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\]

Sum of all the roots\[ = - \dfrac{b}{a}\]

Sum of product of different roots taken two at a time\[ = \dfrac{c}{a}\]

According to the given equation (1), we can say \[a = 1,{\text{ }}b = 3,{\text{ }}c = \dfrac{8}{9},{\text{ }}d = 3\] and \[e = 1\].

Therefore, Sum of all the roots of the given equation (1) is given by \[\alpha + \beta + \dfrac{1}{\alpha } + \dfrac{1}{\beta } = - \dfrac{3}{1} = - 3{\text{ }} \to {\text{(2)}}\]

Also, sum of product of different roots taken two at a time of the given equation (1) is given by

\[\alpha \beta {\text{ + }}\alpha {\text{.}}\dfrac{1}{\alpha }{\text{ + }}\alpha {\text{.}}\dfrac{1}{\beta }{\text{ + }}\beta {\text{.}}\dfrac{1}{\alpha }{\text{ + }}\beta {\text{.}}\dfrac{1}{\beta }{\text{ + }}\dfrac{1}{\alpha }{\text{.}}\dfrac{1}{\beta }{\text{ }} = \dfrac{{\dfrac{8}{9}}}{1} = \dfrac{8}{9}\]

\[

\Rightarrow \alpha \beta {\text{ + }}1{\text{ + }}\alpha {\text{.}}\dfrac{1}{\beta }{\text{ + }}\beta {\text{.}}\dfrac{1}{\alpha }{\text{ + }}1{\text{ + }}\dfrac{1}{\alpha }{\text{.}}\dfrac{1}{\beta }{\text{ }} = \dfrac{8}{9} \Rightarrow \alpha \beta {\text{ + }}\alpha {\text{.}}\dfrac{1}{\beta }{\text{ + }}\beta {\text{.}}\dfrac{1}{\alpha }{\text{ + }}\dfrac{1}{\alpha }{\text{.}}\dfrac{1}{\beta } = \dfrac{8}{9} - 2 = - \dfrac{{10}}{9} \\

\Rightarrow \alpha \left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right){\text{ + }}\dfrac{1}{\alpha }\left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right) = - \dfrac{{10}}{9} \Rightarrow \left( {\alpha + \dfrac{1}{\alpha }} \right)\left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right) = - \dfrac{{10}}{9}{\text{ }} \to {\text{(3)}} \\

\]

Equation (2) can be rearranged as \[\left( {\alpha + \dfrac{1}{\alpha }} \right) = - 3 - \left( {\beta + \dfrac{1}{\beta }} \right)\]

Put the value of \[\left( {\alpha + \dfrac{1}{\alpha }} \right)\] in equation (3), we get

\[\left[ { - 3 - \left( {\beta + \dfrac{1}{\beta }} \right)} \right]\left( {\beta {\text{ + }}\dfrac{1}{\beta }} \right) = - \dfrac{{10}}{9}\]

Let \[\left( {\beta + \dfrac{1}{\beta }} \right) = t\]

\[

\Rightarrow \left[ { - 3 - t} \right]t = - \dfrac{{10}}{9} \Rightarrow - \left( {3 + t} \right)t = - \dfrac{{10}}{9} \Rightarrow \left( {3 + t} \right)t = \dfrac{{10}}{9} \Rightarrow 9{t^2} + 27t - 10 = 0 \\

\Rightarrow 9{t^2} + 27t - 10 = 0 \Rightarrow 9{t^2} - 3t + 30t - 10 = 0 \Rightarrow 3t\left( {3t - 1} \right) + 10\left( {3t - 1} \right) = 0 \\

\Rightarrow \left( {3t - 1} \right)\left( {3t + 10} \right) = 0 \\

\]

i.e., Either \[3t - 1 = 0\] or \[3t + 10 = 0\]

\[

\Rightarrow t = \dfrac{1}{3} \Rightarrow \beta + \dfrac{1}{\beta } = \dfrac{1}{3} \Rightarrow \dfrac{{{\beta ^2} + 1}}{\beta } = \dfrac{1}{3} \Rightarrow 3{\beta ^2} + 3 = \beta \Rightarrow 3{\beta ^2} - \beta + 3 = 0 \Rightarrow \beta = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \times 3 \times 3} }}{{2 \times 3}} \\

\Rightarrow \beta = \dfrac{{1 \pm \sqrt {1 - 36} }}{6} \Rightarrow \beta = \dfrac{{1 \pm \sqrt { - 35} }}{6} \Rightarrow \beta = \dfrac{{1 \pm i\sqrt {35} }}{6} \\

\]

or \[

t = - \dfrac{{10}}{3} \Rightarrow \beta + \dfrac{1}{\beta } = - \dfrac{{10}}{3} \Rightarrow \dfrac{{{\beta ^2} + 1}}{\beta } = - \dfrac{{10}}{3} \Rightarrow 3{\beta ^2} + 3 = - 10\beta \Rightarrow 3{\beta ^2} + 10\beta + 3 = 0 \Rightarrow 3{\beta ^2} + 9\beta + \beta + 3 = 0 \\

\Rightarrow 3\beta \left( {\beta + 3} \right) + 1\left( {\beta + 3} \right) = 0 \Rightarrow \left( {3\beta + 1} \right)\left( {\beta + 3} \right) = 0 \Rightarrow \beta = - \dfrac{1}{3}, - 3 \\

\]

Using equation (2) put the value of \[\beta \], we will get the value of \[\alpha \]

\[ \Rightarrow \alpha = - \dfrac{1}{3}, - 3\] or \[\alpha = \dfrac{{1 \pm i\sqrt {35} }}{6}\]

Therefore, all the roots of the given equation are \[ - \dfrac{1}{3}, - 3,\dfrac{{1 \pm i\sqrt {35} }}{6}\].

Note- These types of problems can be solved by somehow checking for some properties regarding roots of a polynomial and then finding out an appropriate relation between the roots and hence solving further to get the values of these roots.

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