`int_2^8 1/((x-6)^3) dx = -1/(2(x-6)^2)|_2^8 = -1/2(1/(2^2) - 1/(-4)^2)=-1/2(1/4 - 1/16) `

`= -1/2(3/16) = -3/32`

**The integral converges to a value of -3/32**

Since the problem requests to decide if the given integral converges or diverges, then the upper limit must be `oo` and not 8.

Hence, you should evaluate the following limit to check if the integral converges or diverges, such that:

`int_2^oo 1/((x-6)^3)dx = lim_(n->oo) int_2^n 1/((x-6)^3)dx`

You may evaluate the integral `int 1/((x-6)^3)dx` using the following substitution such that:

`x-6 = t => dx = dt`

Changing the variable yields:

`int 1/((x-6)^3)dx= int 1/(t^3)dt`

Using the property of negative power yields:

`int 1/(t^3)dt = int (t^(-3))dt = t^(-3+1)/(-3+1) + c`

`int 1/(t^3)dt = int (t^(-3))dt = -1/(2t^2) + c`

Substituting back `x-6` for t yields:

`int 1/((x-6)^3)dx=-1/(2(x-6)^2) + c `

You may evaluate the definite integral using the fundamental theorem of calculus such that:

`int_2^n 1/((x-6)^3)dx = (-1/(2(x-6)^2))|_2^n`

`int_2^n 1/((x-6)^3)dx =-1/(2(n-6)^2) +1/(2(2-6)^2)`

`int_2^n 1/((x-6)^3)dx = 1/32 - 1/(2(n-6)^2)`

Evaluating the limit yields:

`lim_(n->oo) int_2^n 1/((x-6)^3)dx = lim_(n->oo)(1/32 - 1/(2(n-6)^2))`

`lim_(n->oo) int_2^n 1/((x-6)^3)dx = lim_(n->oo)(1/32) - lim_(n->oo) 1/(2(n-6)^2))`

`lim_(n->oo) int_2^n 1/((x-6)^3)dx =1/32 - (1/2)lim_(n->oo) 1/(n^2(1 - 6/n)^2)`

`lim_(n->oo) int_2^n 1/((x-6)^3)dx = 1/32 - (1/2)*(1/oo)`

`lim_(n->oo) int_2^n 1/((x-6)^3)dx = 1/32 - (1/2)*(0)`

`lim_(n->oo) int_2^n 1/((x-6)^3)dx = 1/32`

**Hence, evaluating the improper integral yields `int_2^oo 1/((x-6)^3)dx = lim_(n->oo) int_2^n 1/((x-6)^3)dx = 1/32` , thus, the intgeral converges.**