Ramanujan 6-8-10 Identity

Let ${a, b, c,}$ and ${d}$ be any numbers such that ${ad=bc.}$ Then

$\displaystyle 64\left\{(a+b+c)^{6}+(b+c+d)^{6}-(c+d+a)^{6}-(d+a+b)^{6}+(a-d)^{6}-(b-c)^{6}\right\}$
$\displaystyle \times\left\{(a+b+c)^{10}+(b+c+d)^{10}-(c+d+a)^{10}-(d+a+b)^{10}+(a-d)^{10}-(b-c)^{10}\right\}$
$\displaystyle =45\left\{(a+b+c)^{8}+(b+c+d)^{8}-(c+d+a)^{8}-(d+a+b)^{8}+(a-d)^{8}-(b-c)^{8}\right\}^{2}$

Let

$\displaystyle \begin{aligned} F_{2 m}(a, b, c, d)=(a+b+c)^{2 m}+(b+c+d)^{2 m}-(c+d+a)^{2 m} & \\ -(d+a+b)^{2 m}+(a-d)^{2 m}-(b-c)^{2 m} &-(b-c)^{2 m} \end{aligned}$

Change of variables $\displaystyle b=a x, c=a y, \text { and } d=a x y$ gives

$\displaystyle F_{2 m}(a, b, c, d)=a^{2 m} f_{2 m}(x, y)$

with ${\begin{aligned} f_{2 m}(x, y)=(1+x+y)^{2 m}+(x+y+x y)^{2 m}-(y+x y+1)^{2 m} \\ &-(x y+1+x)^{2 m}+(1-x y)^{2 m}-(x-y)^{2 m} \end{aligned}}$

The identity now reduces to

$\displaystyle 64 f_{6}(x, y) f_{10}(x, y)=45 f_{8}^{2}(x, y).$

We have the following expressions that prove the identity.

$\displaystyle f_{6}(x, y)=3 x\left(1-x^{2}\right)(x+2)(2 x+1) y\left(1-y^{2}\right)(y+2)(2 y+1)$

$\displaystyle f_{8}(x, y)=8 x\left(1-x^{2}\right)(x+2)(2 x+1)\left(x^{2}+x+1\right) y\left(1-y^{2}\right)(y+2)(2 y+1)\left(y^{2}+y+1\right)$

$\displaystyle f_{10}(x, y)=15 x\left(1-x^{2}\right)(x+2)(2 x+1)\left(x^{2}+x+1\right)^{2} y\left(1-y^{2}\right)(y+2)(2 y+1)\left(y^{2}+y+1\right)^{2}$

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