Published on 29th March 2018, City: Ahmedabad, Country : India.

`A^2+B^2+C^2=D^2` In Ambalal Cuboid Theory - 1,b

If `A=1,B=b` are given where b is any member of `[N,Z,Q,R]`

Then `C=b/(b-1)`

& `D=b/(b-1)+b-1` means `D=C+(b-1)`.

Results: `A^2+B^2+C^2=D^2` become true for any values

of b, where b is any member of `[N,Z,Q,R]`

Proof `D^2-C^2=[b/(b-1)+b-1]^2-[b/(b-1)]^2`

`=[2(b/(b-1))+b-1](b-1)`

`=2b+(b-1)^2`

`=2b+b^2-2b-1`

`=(b)^2-(1)^2`

`=B^2-A^2`

∴ `A^2+B^2+C^2=D^2`

Example 1: Take `B=13`

Now `A=1,B=13,C=13/(13-1)=13/12,D=13/(13-1)+13-1`

`=13/12+12`

`=(13+144)/12`

`=157/12`

Then `A^2+B^2+C^2=D^2`

`(1)^2+(13)^2+(13/12)^2=(157/12)^2`

∴ `(1(144)+(169)(144)+169)/144=(157/12)^2`

∴ `(144+24336+169)/144=24649/144`

Example 2: Take `B=-7`

Let `A=1,B=-7,C=(-7)/(-7-1)=7/8 ∴ D=((-7)/(-7-1))+(-7-1)`

`=(7/8)-8`

`=(7-64)/8=(-57)/8`

Then `A^2+B^2+C^2=D^2`

∴ `(1)^2+(-7)^2+(7/8)^2+(-57/8)^2`

∴ `((1)(64)+(43)(64)+49)/64=3249/64`

∴ `(64+3136+49)/64=3249/64`

Example 3: Take `b=2.5`

∴ `A=1,B=2.5,C=2.5/(25-1)=2.5/1.5=5/3`

`B=5/2`

& `D=5/3+(2.5-1)=5/3+3/2=19/6`

Then `A^2+B^2+C^2=D^2`

`(1)^2+(5/2)^2+(5/3)^2=(19/6)^2`

∴ `((1)(36))/(36)+(25/4)(9/9)+(25/9)(4/4)=361/36`

∴ `(36+225+100)/36=361/36`

Now for satisfaction take `A=1,B=b` where here b is any member of `[N,Z,Q,R]`

`C= b/(b-1)` & `D=C+(b-1)` and put in formula

`A^2+B^2+C^2=D^2` to get successful results.

For any doubt and additional details please contact me via Email : ambalalparmar40@yahoo.com

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