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

`A^2+B^2+C^2+D^2` In Ambalal Cuboid Theory- `ax`

If `A=a,B=a+x` are given where x is any member of `[N,Z,Q]`

Then `C=(a(a+x))/x`

`and D= (a(a+x))/(x) + x` means `D=C+x`

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

value of `x` where `x` is any member of `[N,Z,Q]`


Proof: `D^2-C^2=[(a(a+x))/(x) + x]^2-[(a(a+x))/(x)]^2`

`=[(2a(a+x))/(x) + x][x]`

`=2a(a+x) + x^2`

`=2a^2 + 2ax + x^2`

`=a^2+ (a^2 + 2ax + x^2)`

`=a^2+ (a+x)^2`

`=A^2+B^2`

`∴ A^2+B^2+C^2=D^2`

Example 1:

Take `x=7` and `a=31`.

Now `A=31, B=31+7 = 38, C=(31(31+7))/7 = 1178/7, D=C+1 = 1178/7 + 7 = (1178+ 49)/7 = 1227/7`

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

Then `(31)^2+(38)^2+(1178/7)^2=(1227/7)^2`

`∴ (961(49))/49 + 1444 (49/49) + 1387684/49 = 1505529/49`

`∴ 47089 + 70756 + 1387684 = 1505529.`

Example 2:

Take `x=-5` and `A=13`
Now `A=13, B= 13 +(-5) = 8, C=(13(13-5))/-5 = 104/-5, D = C+x = 104/-5 + -5 = (104+25)/-5 =129/-5`

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

Then `(13)^2+(8)^2+(104/-5)^2=(129/-5)^2`

`∴ (169(25))/((25)) + (64(25))/((25)) + 10816/25 = 16641/25`

`∴ 4225 + 1600 + 10816 = 16641.`

Example 3:

Take `x=0.7` and `a=2.3`
Now `A=2.3, B= 2.3 + 0.7 = 3, C=2.3((2.3+0.7))/0.7 = 6.9/0.7, D = C+x = 6.9/0.7 + 0.7 = (6.9 + 0.49 )/ 0.7 = 7.39/0.7`

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

Then `(2.3)^2+(3)^2+(6.9/0.7)^2 = (7.39 / 0.7)^2`

`∴ (5.29(0.49))/((0.49)) + (9 (0.49))/((0.49)) + 47.61/((0.49)) = 54.6121/((0.49))`

`∴ 2.5921 + 4.41 + 47.61 = 54.6121.`



Now for your satisfaction, take `A=a, B = a + x` where x is any member of `[N,Z,Q]`

`C=(a(a + x))/x, and D= c+ x ` and put in formula `A^2+B^2+C^2=D^2` to get successful result.

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

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