Răspuns:
x = 0 ; x = 1
Explicație pas cu pas:
I (x+1)³ - (x-2)³ I = 9 <=>
I (x+1)(x²+2x+1) - (x-2)(x²-4x+4) I = 9 <=>
I (x³+2x²+x+x²+2x+1) - (x³-4x²+4x-2x²+8x-8) I = 9 <=>
I x³+3x²+3x+1 - x³+6x²-12x+8 I = 9 <=>
I 9x²-9x+9 I = 9 <=>
9· I x²-x+1 I = 9 I:9 <=>
I x²-x+1 I = 1 <=>
I x²-x I = 0 <=>
I x·(x-1) I = 0 => x = 0 ; x = 1
