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Find all solutions of x(x^3 - 1) - 6(x^2 + x + 1) = 0.

 Dec 9, 2019
 #1
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+1

x*(x-1)(x^2+x+1)=6(x^2+x+1)

Notice that x^2+x+1 is on both sides so divide by it 

x*(x-1)=6

x^2-x-6=0 

(x-3)(x+2)=0

x= 3

x=-2 

For real numbers! 

 Dec 9, 2019
 #2
avatar+109241 
+1

x ( x^3 - 1) - 6(x^2 + x + 1)  =  0

 

Factor  (x^3 - 1)  as  a difference of cubes

 

x  ( x - 1) (x^2 + x + 1)  - 6( x^2 + x + 1)   =   0

 

(x^2  + x + 1)  [ x(x - 1) - 6  ]  = 0

 

(x^2 + x + 1) [  x^2 - x - 6 ] =  0

 

(x^2 + x + 1)  ( x - 3) ( x + 2)  =  0

 

The first factor doesn't  produce any real number roots  when set to 0 and solved

 

The  real roots   are  found  by setting the other two factors to 0 and solving for  x ⇒    x   =3   and  x  = -2

 

 

cool cool cool

 Dec 9, 2019

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