Odpowiedź :
[tex]W(x) = x^{3}+(a^{3}-1)x^{2}+(2a^{2}+4a+23)x - 15\\\\x_1 = 1\\\\W(1) = 0\\\\1^{3}+(a^{3}-1)\cdot1^{2} + (2a^{2}+4a+23)\cdot1 - 15 = 0\\\\1 +(a^{3}-1)\cdot1 + (2a^{2}+4a+23)\cdot1 -15 = 0\\\\1+a^{3}-1+2a^{2}+4a+23 - 15 = 0\\\\a^{3}+2a^{2}+4a + 8 = 0\\\\a^{2}(a+2) + 4(a+2) = 0\\\\(a+2)(a^{2}+4) = 0\\\\a+2 = 0 \ \vee \ a^{2}+4 = 0\\\\a = -2, \ a^{2}\neq -4\\\\a = -2[/tex]
[tex]W(x) = x^{3}+(a^{3}-1)x^{2} + (2a^{2}+4a+23)x -15\\\\W(x) = x^{3}+[(-2)^{3}-1)]x^{2} + [2(-2)^{2}+4(-2)+23]x - 15\\\\W(x) = x^{3}+(-8-1)x^{2} + (8-8+23)x - 15\\\\W(x) = x^{3} -9x^{2}+23x - 15[/tex]
[tex]W(x) = x^{3}-x^{2}-8x^{2}+8x+15x-15\\\\W(x) = x^{2}(x-1) -8x(x-1) +15(x-1)\\\\W(x) = (x-1)(x^{2}-8x+15)\\\\W(x) = (x-1)(x^{2}-5x-3x+15)\\\\W(x) = (x-1)[x(x-5)-3(x-5)]\\\\W(x) = (x-1)(x-3)(x-5)\\\\W(x) = 0\\\\(x-1)(x-3)(x-5) = 0\\\\x-1=0 \ \vee \ x-3 = 0 \ \vee \ x-5 = 0\\\\x = 1 \ \vee \ x = 3 \ \vee \ x = 5[/tex]
[tex]x \in \{1, 3, 5\}[/tex]