Question 975311
<pre>
 If (x-a)(x-b)=1 and a-b+5=0
Then
(x-a)^3-1/(x-a)^3 = ?

71 +- 13V29

Let x-a = u.  Then x = a+u

Making those substitutions the problem becomes:

If u(a+u-b)=1 and a-b+5=0

Then
u^3-1/u^3 = ?

Since a-b+5=0, then a = b-5

Substitute in

u(a+u-b)=1

u(b-5+u-b) = 1

u(-5+u) = 1

-5u+uČ = 1

uČ-5u-1 = 0

To avoid a conflict of letters, we use capital letters 
in the quadratic formula

{{{u}}}{{{""=""}}}{{{(-B +- sqrt( B^2-4AC ))/(2A) }}}


{{{u}}}{{{""=""}}}{{{(-(-5) +- sqrt((-5)^2-4(1)(-1) ))/(2(1)) }}}

{{{u}}}{{{""=""}}}{{{(5 +- sqrt(25+4 ))/2 }}}

{{{u}}}{{{""=""}}}{{{(5 +- sqrt(29))/2 }}}

We want to find {{{u^3-1/u^3}}}: 

We factor as the difference of two cubes:

(1)  {{{u^3-1/u^3}}}{{{""=""}}}{{{(u-1/u)(u^2+u(1/u)+1/u^2)}}}{{{""=""}}}{{{u^2+1+1/u^2}}}

We need u<sup>2</sup>, {{{1/u}}}, {{{u-1/u}}}, and {{{1/u^2}}} 

{{{u^2}}}{{{""=""}}}{{{((5 +- sqrt(29))/2)^2 }}}

Let's first use the + sign, {{{u}}}{{{""=""}}}{{{(5 + sqrt(29))/2 }}}

{{{u^2}}}{{{""=""}}}{{{(25 + 10sqrt(29)+29)/4 }}}

{{{u^2}}}{{{""=""}}}{{{(54 + 10sqrt(29))/4 }}}

{{{u^2}}}{{{""=""}}}{{{(2(27 + 5sqrt(29)))/4 }}}

{{{u^2}}}{{{""=""}}}{{{(27 + 5sqrt(29))/2 }}}

{{{1/u}}}{{{""=""}}}{{{2/(5 + sqrt(29)) }}}

Rationalizing the denominator:

{{{1/u}}}{{{""=""}}}{{{2/(5 + sqrt(29))}}}{{{""*""}}}{{{(5-sqrt(29))/(5 - sqrt(29)) }}}{{{""=""}}}{{{(2(5 - sqrt(29)))/(5^2-29) }}}{{{""=""}}}{{{(2(5 - sqrt(29)))/(-4) }}}{{{( 5-sqrt(29))/(-2) }}}{{{""=""}}}{{{(sqrt(29)-5)/2}}}

{{{u-1/u}}}{{{""=""}}}{{{(5 + sqrt(29))/2 - (sqrt(29)-5)/2}}}{{{""=""}}}{{{((5 + sqrt(29))- (sqrt(29)-5))/2}}}{{{""=""}}}{{{(5 + sqrt(29)- sqrt(29)+5)/2 }}}{{{""=""}}}{{{10/2}}}{{{""=""}}}{{{5}}}

{{{1/u^2}}}{{{""=""}}}{{{(1/u)^2}}}{{{""=""}}}{{{((sqrt(29)-5)/2)^2}}}{{{""=""}}}{{{(29-10sqrt(29)+25)/4}}}{{{""=""}}}{{{(54-10sqrt(29))/4}}}{{{""=""}}}{{{(2(27-5sqrt(29)))/4}}}{{{""=""}}}{{{(27-5sqrt(29))/2}}}

Going back to equation (1)

{{{u^3-1/u^3}}}{{{""=""}}}{{{(u-1/u)(u^2+1+1/u^2)}}}{{{""=""}}}

{{{(5)( (27 + 5sqrt(29))/2+1+(27-5sqrt(29))/2)}}}

{{{""=""}}}{{{5((27 + 5sqrt(29))/2+2/2+(27-5sqrt(29))/2)


}}}{{{""=""}}}{{{5(56/2)}}}{{{""=""}}}{{{5(28)}}}{{{""=""}}}{{{140}}}

If we do the same with {{{u}}}{{{""=""}}}{{{(5 - sqrt(29))/2 }}},
all the signs of the terms in {{{sqrt(29)}}} will reverse their signs,
and we'll still get 140.

Answer: 140

Edwin</pre>