Let ( u = x^3 + 1 ).
Then, the equation becomes:
( u^2 - 10u + 9 = 0 )
This is a quadratic equation that can be factored as:
( (u-9)(u-1) = 0 )
Setting each factor to zero:
( u - 9 = 0 ) or ( u - 1 = 0 )
( u = 9 ) or ( u = 1 )
Now substitute back in ( x^3 + 1 ) for ( u ):
For ( u = 9 ):
( x^3 + 1 = 9 )
( x^3 = 8 )
( x = 2 )
For ( u = 1 ):
( x^3 + 1 = 1 )
( x^3 = 0 )
( x = 0 )
Therefore, the solutions to the original equation are ( x = 2 ) and ( x = 0 ).
Let ( u = x^3 + 1 ).
Then, the equation becomes:
( u^2 - 10u + 9 = 0 )
This is a quadratic equation that can be factored as:
( (u-9)(u-1) = 0 )
Setting each factor to zero:
( u - 9 = 0 ) or ( u - 1 = 0 )
( u = 9 ) or ( u = 1 )
Now substitute back in ( x^3 + 1 ) for ( u ):
For ( u = 9 ):
( x^3 + 1 = 9 )
( x^3 = 8 )
( x = 2 )
For ( u = 1 ):
( x^3 + 1 = 1 )
( x^3 = 0 )
( x = 0 )
Therefore, the solutions to the original equation are ( x = 2 ) and ( x = 0 ).