A Reduction Using Ø

How does the mess {[\/¯(16ز - 3) - Ø\/¯(16 - 3ز)] / 8Ø}² reduce elegantly to
1 / 64?????

This reduction shows the power of the Ø ratio to combine and divide into itself.

I could simply have used repeating decimals and avoided this complicated reduction. However, as I have remarked before, writing \/¯(16ز - 3) is stating a PRECISE QUANTITY.
Writing 6.236067979......  is sloppy. It is a repeating decimal and is only an approximation. It inherently results in loss of information. It is inelegant.
In mathematics it is always better to be precise than lazy.
When this reduction is performed on a calculator, round-off error and the imprecision of the repeating decimal yields accuracy to 7 decimal places. That is pretty good, but not perfect! Using repeating decimals yields 1 / 63.99999994 instead of 1 / 64.

Here is the reduction:
{[\/¯(16ز - 3) - Ø\/¯(16 - 3ز)] / 8Ø}² =
(16ز - 3) + ز(16 - 3ز) - 2Ø\/¯[(16ز - 3) *(16 - 3ز)]   / 64ز.

Now we will break down  (16ز - 3) *(16 - 3ز) .
(16ز - 3) *(16 - 3ز) = 256ز - 48Ø^4 - 48 + 9ز
                                   = -48Ø^4 + 265ز - 48

Now note that Ø^4 = س + ز  and  س = ز + Ø   so that Ø^4 = 2ز + Ø.

(16ز - 3) *(16 - 3ز) = -96ز - 48Ø + 265ز - 48
                                    = 169ز - 48(Ø + 1)

Now note that ز = Ø + 1

                                    = 169ز - 48ز
                                    = 121ز

So \/¯[(16ز - 3) *(16 - 3ز)] = 11Ø.

Now we can write:
(16ز - 3) + ز(16 - 3ز) - 2Ø\/¯[(16ز - 3) *(16 - 3ز)]   / 64ز =
(16ز - 3) + ز(16 - 3ز) - 2Ø(11Ø) =
16ز -3 + 16ز - 3Ø^4 -22ز =
32ز -3 -3Ø^4 - 22ز =
-3Ø^4 + 10ز - 3 =
-3(2ز + Ø) + 10ز -3 =
4ز - 3Ø - 3=
4ز -3(Ø + 1) =
4ز - 3ز =
ز  !!!

Now we have to divide this by 64ز, and we get
1 / 64.

At all times we have used exact mensuration.
We have not approximated, we have not yielded  one iota of error  at any step in our calculations.
That is the power of the division into mean and extreme ratio.

Special characters:
 \/¯ ° ¹ ² ³ × ½ ¼ Ø  \/¯(ز + 1)