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Find all real numbers x such that (3x - 27)^3 + (27x - 3)^3 = (3x + 27x - 30).

 Aug 18, 2022
 #1
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(3x - 27)^3  + (27x - 3)^3  =   30x  - 30

 

Factor left side as a sum of cubes.....

 

[ (3x - 27 + 27x - 3) ]  [  (3x - 27)^2 - (3x - 27)(27x - 3) + ( 27x - 3)^2 ]  =  30x - 30

 

( 30x - 30)  [ (3(x - 9))^2 - ( 3 (x - 9) * 3(9x - 1) ) + (3 (9x - 1))^2 ] = 30x -30

 

(30x - 30)  [ 9 ( x^2 - 18x + 81)  - ( 9(9x^2 - 82x + 9)) + (9 ( 81x^2 - 18x + 1))  = 30x - 30

 

9 (30x - 30) [ x^2 - 18x + 81 - 9x^2 + 82x - 9 + 81x^2 - 18x + 1 ]  = 30x - 30

 

270 ( x - 1)  [ 73x^2 + 46x + 73 ]  =  30 (x - 1)

 

(270 - 30) ( x - 1)  [ 73x^2 + 46x^2 + 73 ]  =  0

 

(240) ( x - 1) [ 73x^2 + 46x + 73 ]  =  0           divide through by 240

 

(x - 1)  (73x^2 + 46x + 73 ]   =  0

 

The discriminant  of   73x^2 + 46x^2 + 73  =  [ 46^2 - 4 (73)^2 ]  = negative....so  this factor has no real  roots

 

The  only real solution is

 

(x  -1 )  = 0

 

x  = 1

 

 

cool cool cool

 Aug 18, 2022

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