Try me!

We can make a equation having its roots as $\frac{1+\alpha}{1-\alpha},\frac{1+\beta}{1-\beta}$ and $\frac{1+\gamma}{1-\gamma}$ .

Let $y=\frac{1+x}{1-x}$ , then $x=\frac{y-1}{y+1}$ .

Placing $x=\frac{y-1}{y+1}$ in $x^3-x-1=0$ .

$\left(\frac{y-1}{y+1}\right)^3-\left(\frac{y-1}{y+1}\right)+1=0 \\ (y-1)^3-(y-1)(y+1)^2-(y+1)^3=0 \\ y^3+7y^2-y+1=0$

So, $y^3+7y^2-y+1=0$ has $\frac{1+\alpha}{1-\alpha},\frac{1+\beta}{1-\beta}$ and $\frac{1+\gamma}{1-\gamma}$ as its roots.

Therefore, to find sum of second power of roots, we use Newton's sum method and our answer comes as $\boxed{51}$ .

Nice question Akshat! It requires witty substitution of variables. Firstly, as $\alpha$ , $\beta$ and $\gamma$ are roots of the equation $x^3-x-1=0$ , they must satisfy the given equation also.

Placing value of $\alpha$ (similar way for other variables),

$\alpha^3-\alpha-1=0$

$\alpha+1=\alpha^3$

And also,

$\alpha(\alpha^2-1)-1=0$

$(\alpha+1)(\alpha-1)=\frac{1}{\alpha}$

$\alpha-1=\frac{1}{\alpha^4}$

Hence the question can be reframed as,

$\displaystyle \sum_{\alpha,\beta,\gamma} \left( \frac{1+\alpha}{1-\alpha}\right)^2= \displaystyle \sum_{\alpha,\beta,\gamma} \left( \frac{\alpha+1}{\alpha-1}\right)^2=\left( \frac{\alpha+1}{\alpha-1}\right)^2+\left( \frac{\beta+1}{\beta-1}\right)^2+\left( \frac{\gamma+1}{\gamma-1}\right)^2=\left(\frac{\alpha^3}{1/\alpha^4}\right)^2+\left(\frac{\beta^3}{1/\beta^4}\right)^2+\left(\frac{\gamma^3}{1/\gamma^4}\right)^2$

$=\alpha^{14}+\beta^{14}+\gamma^{14}$

This required sum can be calculated by Newton's sum method .

After tedious calculations you get,

$\alpha^{14}+\beta^{14}+\gamma^{14}=\boxed{51}$

Since $(1-x)(1+2x+2x^2) + x+1 = 2(1+x-x^3)$ , we have $\tfrac{1+\alpha}{1-\alpha} \,=\, -(1+2\alpha+2\alpha^2) \qquad \tfrac{1+\beta}{1-\beta} \,=\, -(1+2\beta+2\beta^2) \qquad \tfrac{1+\gamma}{1-\gamma} \,=\, -(1+2\gamma+2\gamma^2)$ and so $\sum_\alpha \left(\tfrac{1+\alpha}{1-\alpha}\right)^2 \; = \; \sum_\alpha (1 + 2\alpha+2\alpha^2)^2 \; =\; S_0 + 4S_1 + 8S_2 + 8S_3 + 4S_4$ where $S_n = \alpha^n + \beta^n + \gamma^n$ . Using Newton's formulae, the answer is $3+0+16+24+8 = \boxed{51}$ .

Alternatively, the substitution $x = \frac{\sqrt{y}-1}{\sqrt{y}+1}$ gives the cubic $y^3 - 51y^2 - 13y - 1 = 0$ as the equation with roots $\left(\frac{1+\alpha}{1-\alpha}\right)^2$ , $\left(\frac{1+\beta}{1-\beta}\right)^2$ and $\left(\frac{1+\gamma}{1-\gamma}\right)^2$ , again yielding $51$ as the answer.

For variety, here's one more solution:

$\displaystyle \sum_{n=\{\alpha,\beta,\gamma\}}\left(\dfrac{1+n}{1-n}\right)^2\\ =\displaystyle \sum_{n=\{\alpha,\beta,\gamma\}}\left(\dfrac{n+1}{n-1}\right)^2\\ =\left(\dfrac{\alpha+1}{\alpha-1}\right)^2 + \left(\dfrac{\beta+1}{\beta-1}\right)^2 + \left(\dfrac{\gamma+1}{\gamma-1}\right)^2$

From Vieta's formula, we know that

$\alpha + \beta + \gamma = 0\\ \alpha\beta + \alpha\gamma + \beta\gamma = -1\\ \alpha\beta\gamma = 1$

Now, to simplify our calculations, we substitute $\alpha = a + 1$ , $\beta = b+1$ and $\gamma = c+1$ into the expression and the equations

The expression becomes:

$\left(\dfrac{a+2}{a}\right)^2 + \left(\dfrac{b+2}{b}\right)^2 + \left(\dfrac{c+2}{c}\right)^2\\ =\dfrac{a^2+4a+4}{a^2} + \dfrac{b^2+4b+4}{b^2} + \dfrac{c^2 + 4c +4}{c^2}\\ =\dfrac{b^2c^2(a^2+4a+4) + a^2c^2(b^2 + 4b + 4) + a^2b^2(c^2+4c+4)}{a^2b^2c^2}\\ =\dfrac{3a^2b^2c^2 + 4ab^2c^2 + 4a^2bc^2 + 4a^2b^2c + 4a^2b^2 + 4a^2c^2 + 4b^2c^2}{a^2b^2c^2}\\ =\dfrac{3(abc)^2 + 4abc(ab+ac+bc) + 4(a^2b^2+a^2c^2+b^2c^2)}{(abc)^2}$

The equations become (simplify yourself, I'm too lazy to type out all the steps):

$a + 1 + b + 1 + c + 1 = 0 \implies a+b+c = -3\\ (a+1)(b+1) + (a+1)(c+1) + (b+1)(c+1) = -1 \implies ab+ac+bc = 2\\ (a+1)(b+1)(c+1) = 1 \implies abc=1$

As an additional step, calculate the value of $a^2b^2 + a^2c^2 + b^2c^2$ :

$(ab+ac+bc)^2 = a^2b^2 + a^2c^2 + b^2c^2 + 2(a^2bc + ab^2c + abc^2)\\ a^2b^2 + a^2c^2 + b^2c^2 = (ab+ac+bc)^2 - 2abc(a+b+c) = 2^2 - 2(1)(-3) = 10$

Substitute all the values into the expression:

$\dfrac{3(abc)^2 + 4abc(ab+ac+bc) + 4(a^2b^2+a^2c^2+b^2c^2)}{(abc)^2}\\ =\dfrac{3(1)^2 + 4(1)(2) + 4(10)}{1^2}\\ =3 + 8 + 40\\ =\boxed{51}$

Relevant wiki: Newton's Identities

$For\quad those\quad who\quad don't\quad know\quad newton\quad sums(as\quad it\quad is\quad not\quad a\quad part\quad of\quad JEE\quad syllabus)\\ First\quad notice\quad that\quad we\quad can\quad write\\ \frac { \alpha +1 }{ \alpha -1 } \quad as\quad 1+\frac { 2 }{ \alpha -1 } \\ so\quad the\quad required\quad sum\quad becomes\\ \sum _{ \alpha ,\beta ,\gamma }^{ }{ { \left( 1+\frac { 2 }{ \alpha-1 } \right) }^{ 2 } } =3+4\left( \sum { \frac { 1 }{ { (\alpha -1) }^{ 2 } } } \right) +4\left( \sum { \frac { 1 }{ (\alpha -1) } } \right) \\ Now,\\ { x }^{ 3 }-{ x }^{ 2 }-1=(x-\alpha )(x-\beta )(x-\gamma )\\ taking\quad log\quad on\quad both\quad sides\quad and\quad differentiating\\ \frac { 3{ x }^{ 2 }-1 }{ 1+{ x }^{ 2 }-{ x }^{ 3 } } =\sum { \frac { 1 }{ (\alpha -x) } } \\ Putting\quad x=1\quad we\quad get\quad \\ \sum { \frac { 1 }{ \alpha -1 } } =2\\ Now\quad in\quad the\quad original\quad equation\quad doing\quad the\quad transforations\quad x\rightarrow x+1\quad and\quad after\quad that\quad x\rightarrow \frac { 1 }{ x } \\ -{ x }^{ 3 }+2{ x }^{ 2 }+3x+1=\prod { \left( x+\frac { 1 }{ \alpha -1 } \right) } \\ now\quad let\quad a=\frac { 1 }{ \alpha -1 } ,\quad b=\frac { 1 }{ \beta -1 } ,\quad c=\frac { 1 }{ \gamma -1 } \\ \sum { \frac { 1 }{ { \left( \alpha -1 \right) }^{ 2 } } } =\sum { { a }^{ 2 } } \\ which\quad after\quad calculating\quad using\quad Vieta's\quad we\quad get\\ \sum { \frac { 1 }{ { \left( \alpha -1 \right) }^{ 2 } } } =\sum { { a }^{ 2 } } =10\\ hence\quad our\quad original\quad sum\quad is\\ 4\times 10+4\times 2+3=40+8+3=51$

Replace x by a-1/a+1 where a = square root of x to get the equation whose roots are (1+x/1-x)^2 . Simplify and see sum of roots