Solve the equation x 6 − 2 x 5 + 3 x 4 − 3 x 2 + 2 x − 1 = 0Let's divide both sides of the equation by x 3 ≠ 0 (as x = 0 is obviously not a solution, we can consider x ≠ 0). Then we have x 3 − 2 x 2 + 3 x − 3 ⋅ 1 x + 2 ⋅ 1 x 2 − 1 x 3 = 0 ( x 3 − 1 x 3 ) − 2 ( x 2 − 1 x 2 ) + 3 ( x − 1 x ) = 0What do we do now? If we say y = x − 1 x , we won't be able to express ( x 2 − 1 x 2 ) in terms of y because of the minus sign as ( a − b ) 2 = a 2 − 2 a b + b 2 . On the other side, y 3 = x 3 − 1 x 3 − 3 ( x − 1 x ) x 3 − 1 x 3 = y 3 + 3 yI think this is a traditional issue when solving reciprocal equations, but I can't figure out how to deal with it.

Question

Solve the equation       x    6    −  2      x    5    +  3      x    4    −  3      x    2    +  2  x  −  1  =  0Let&#039;s divide both sides of the equation by       x    3    ≠  0 (as   x  =  0 is obviously not a solution, we can consider   x  ≠  0). Then we have      x    3    −  2      x    2    +  3  x  −  3  ⋅            1      x        +  2  ⋅            1              x        2              −            1              x        3              =  0        (          x      3        −                  1                  x          3                      )    −  2      (          x      2        −                  1                  x          2                      )    +  3      (    x    −                  1        x              )    =  0What do we do now? If we say   y  =  x  −            1      x      , we won&#039;t be able to express       (          x      2        −                  1                  x          2                      )   in terms of   y because of the minus sign as   (  a  −  b      )    2    =      a    2    −  2  a  b      +          b      2      . On the other side,      y    3    =      x    3    −            1              x        3              −  3      (    x    −                  1        x              )          x    3    −            1              x        3              =      y    3    +  3  yI think this is a traditional issue when solving reciprocal equations, but I can&#039;t figure out how to deal with it.

Chaya Proctor · Accepted Answer

Noticing that   x  =  1 and   x  =  −  1 are roots and by long division, you&#039;ll get       x    6    −  2      x    5    +  3      x    4    −  3      x    2    +  2  x  −  1  =  (  x  −  1  )  (  x  +  1  )  (      x    4    −  2      x    3    +  4      x    2    −  2  x  +  1  )Considering       x    4    −  2      x    3    +  4      x    2    −  2  x  +  1  =  0, you now divide by       x    2   and you&#039;ll get       x    2    +      1          x      2        −  2  (  x  +      1    x    )  +  4  =  0 which, using the substitution   y  =  x  +      1    x  , gives       y    2    −  2  y  +  2  =  0.I believe you can finish this from here :)

Poklekom5zc · Accepted Answer

(          x      3        −                  1                  x          3                      )    −  2      (          x      2        −                  1                  x          2                      )    +  3      (    x    −                  1        x              )    =  0There are two roots   ±  1 associated with the factor of   x  −  1      /    x. To find the other roots, divide by   x  −  1      /    x to get      x    2    +  1  +            1              x        2              −  2      (    x    +                  1        x              )    +  3  =  0Noticing that   (  x  +  1      /    x      )    2    =      x    2    +  2  +  1      /        x    2   we have  (  x  +  1      /    x      )    2    −  2  (  x  +  1      /    x  )  +  2  =  0  ,This quadratic has solutions   x  +  1      /    x  =  1  ±  i. Then you need to solve       x    2    −  (  1  ±  i  )  x  +  1  =  0 to get the final roots.

Solve the equation x^6-2x^5+3x^4-3x^2+2x-1=0

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