Roots of a Cubic

Not sure whether there's a better solution without bashing like I'll be: First, notice that by Vieta's Formula , we get the following relations between the $3$ roots: $\begin{cases} \alpha + \beta + \gamma = 6\\ \alpha \beta + \alpha \gamma + \beta \gamma = 10 \\ \alpha \beta \gamma = -1 \end{cases}$ Here I'll be manipulating the above 3 equations (labelled [1], [2], [3] in order of up to down) to give the following expressions to solve the following numerator and denominator expressions. (It is more intuitive to come up with the expressions for denominator and numerator first but I chose to present it this way so that solution will be cleaner.) $\begin{cases} \alpha^2 + \beta^2 + \gamma^2 = [1]^2 - 2[2] = 16 \\ \alpha^2 \beta^2+ \alpha^2 \gamma^2 + \beta^2 \gamma^2 = [2]^2 - 2[3][1] = 112 \end{cases}$ Simply converting the RHS expression into 1 fraction (with a common denominator) we get:

Numerator $\begin{array} {lcl} & = &(\beta^2 + \gamma^2)(\alpha^2 + \gamma^2) + (\alpha^2 + \beta^2)(\alpha^2 + \gamma^2) + (\alpha^2 + \beta^2)(\beta^2 +\gamma^2) \\ & = & \alpha^4 + \beta^4 +\gamma^4 + 3(\alpha^2 \beta^2+ \alpha^2 \gamma^2 + \beta^2 \gamma^2) \\ & = & (\alpha^2 + \beta^2 + \gamma^2)^2 + (\alpha^2 \beta^2+ \alpha^2 \gamma^2 + \beta^2 \gamma^2) \\ & = & 16^2 + 112 \\ & = & 368 \end{array}$

Denominator $\begin{array} {lcl} & = & (\alpha^2 + \beta^2)(\alpha^2 + \gamma^2)(\beta^2 +\gamma^2) \\ & = & (16 - \alpha^2)(16- \beta^2)(16 - \gamma^2) \\ & = & 16^3 - (\alpha \beta \gamma)^2 - 16^2(\alpha^2 + \beta^2 + \gamma^2) + 16(\alpha^2 \beta^2+ \alpha^2 \gamma^2 + \beta^2 \gamma^2) \\ & = & 1791 \end{array}$

Therefore, $B - A = 1791 - 368 = \boxed{1423}.$ . We're done. :)

Intuition Basically when faced with such question - finding a value of an expression in terms of some roots, the most straightforward method is to use Vieta's to derive the first few relations between the roots. Then, we manipulate the equations (from Vieta's) as well as the given expression to get the value we desire.

lets review vieta's $\begin{cases} \alpha + \beta + \gamma = 6\\ \alpha \beta + \alpha \gamma + \beta \gamma = 10 \\ \alpha \beta \gamma = -1 \end{cases}$ and lets compute some important terms $\alpha^2 + \beta^2 + \gamma^2 =(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma +\gamma\alpha)=6^2-20=16$ $\alpha^2\beta^2+ \alpha^2 \gamma^2 + \beta^2 \gamma^2 =(\alpha\beta+\beta\gamma +\gamma\alpha)^2-2 \alpha \beta \gamma (\alpha + \beta + \gamma )=10^2+12=112$ now, the Numerator $\begin{array} {lcl} & = &(\beta^2 + \gamma^2)(\alpha^2 + \gamma^2) + (\alpha^2 + \beta^2)(\alpha^2 + \gamma^2) + (\alpha^2 + \beta^2)(\beta^2 +\gamma^2) \\ & = & \alpha^4 + \beta^4 +\gamma^4 + 3(\alpha^2 \beta^2+ \alpha^2 \gamma^2 + \beta^2 \gamma^2) \\ & = & (\alpha^2 + \beta^2 + \gamma^2)^2 + (\alpha^2 \beta^2+ \alpha^2 \gamma^2 + \beta^2 \gamma^2) \\ & = & 16^2 + 112 \\ & = & 368 \end{array}$ the denominator $(\alpha ^2+\beta^2)(\beta^2+\gamma^2)(\gamma^2+\alpha^2)$ by Daniel Liu 's identity $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ so, $\begin{array}{c}= (\alpha ^2+\beta^2)(\beta^2+\gamma^2)(\gamma^2+\alpha^2)\\=(\alpha^2 + \beta^2 + \gamma^2 )(\alpha^2\beta^2+ \alpha^2 \gamma^2 + \beta^2 \gamma^2)-\alpha^2\beta^2\gamma^2\\=16\times112-1\\=1792-1\\=1791\end{array}$ so the whole expression is $\dfrac{368}{1791}$ and $1791-368=\boxed{1423}$

For easy typing, let $\alpha=a,\;\beta=b,\;\gamma=c$

From Vieta's formula, we know that:

$\color{#3D99F6}{a+b+c=6}\\ \color{#D61F06}{ab+ac+bc=10}\\ \color{#EC7300}{abc=-1}$

To make the sum a bit simpler to calculate, find $\color{#20A900}{a^2+b^2+c^2}$ using Newton's sums:

$\color{#20A900}{a^2+b^2+c^2}=(\color{#3D99F6}{a+b+c})^2-2(\color{#D61F06}{ab+ac+bc})=(\color{#3D99F6}{6})^2-2(\color{#D61F06}{10})=\color{#20A900}{16}$

Our required sum is:

$\dfrac{1}{\color{#20A900}{a^2+b^2}}+\dfrac{1}{\color{#20A900}{b^2+c^2}}+\dfrac{1}{\color{#20A900}{a^2+c^2}}\\ =\dfrac{1}{\color{#20A900}{16-c^2}}+\dfrac{1}{\color{#20A900}{16-a^2}}+\dfrac{1}{\color{#20A900}{16-b^2}}\\ =\dfrac{(16-a^2)(16-b^2)+(16-b^2)(16-c^2)+(16-a^2)(16-c^2)}{(16-a^2)(16-b^2)(16-c^2)}\\ =\dfrac{3(16^2)-2(16)(\color{#20A900}{a^2+b^2+c^2})+\color{#69047E}{a^2b^2+a^2c^2+b^2c^2}}{16^3-16^2(\color{#20A900}{a^2+b^2+c^2})+16(\color{#69047E}{a^2b^2+a^2c^2+b^2c^2})-a^2b^2c^2}\\ =\dfrac{3(16^2)-2(16)(\color{#20A900}{16})+\color{#69047E}{a^2b^2+a^2c^2+b^2c^2}}{16^3-16^2(\color{#20A900}{16})+16(\color{#69047E}{a^2b^2+a^2c^2+b^2c^2})-(\color{#EC7300}{abc})^2}\\ =\dfrac{3(16^2)-2(16^2)+\color{#69047E}{a^2b^2+a^2c^2+b^2c^2}}{16^3-16^3+16(\color{#69047E}{a^2b^2+a^2c^2+b^2c^2})-(\color{#EC7300}{-1})^2}\\ =\dfrac{16^2+\color{#69047E}{a^2b^2+a^2c^2+b^2c^2}}{16(\color{#69047E}{a^2b^2+a^2c^2+b^2c^2})-1}$

Now, all we need to do is find this:

$(\color{#D61F06}{ab+ac+bc})^2=\color{#69047E}{a^2b^2+a^2c^2+b^2c^2}+2(a^2bc+ab^2c+abc^2)\\ \color{#69047E}{a^2b^2+a^2c^2+b^2c^2}=(\color{#D61F06}{ab+ac+bc})^2-2\color{#EC7300}{abc}(\color{#3D99F6}{a+b+c})=(\color{#D61F06}{10})^2-2(\color{#EC7300}{-1})(\color{#3D99F6}{6})=\color{#69047E}{112}$

Substitute this value in:

$\dfrac{16^2+\color{#69047E}{a^2b^2+a^2c^2+b^2c^2}}{16(\color{#69047E}{a^2b^2+a^2c^2+b^2c^2})-1}\\ =\dfrac{16^2+\color{#69047E}{112}}{16(\color{#69047E}{112})-1}\\ =\dfrac{256+112}{1792-1}\\ =\dfrac{368}{1791}$

$\implies A=368,\; B=1791,\; B-A=1791-368=\boxed{1423}$