Cubes and cubes

Relevant wiki: Algebraic Identities

Let $\alpha = x^2+x-2$ , $\beta = 2x^2-x-1$ and $\gamma = -3(x^2-1)$ .

Then, we note that: $\begin{aligned} \alpha + \beta + \gamma & = x^2+x-2 + 2x^2-x-1 -3x^2+3 = 0 \end{aligned}$

For $\alpha + \beta + \gamma = 0$ $\implies \alpha^3 + \beta^3 + \gamma^3 = 3 \alpha \beta \gamma$ . Since $\alpha^3 + \beta^3 + \gamma^3 = 0$ $\implies 3 \alpha \beta \gamma = 0$ . Therefore, we have:

$\begin{aligned} \alpha \beta \gamma & = 0 \\ (x^2+x-2)(2x^2-x-1)(3-3x^2) & = 0 \\ (x^2+x-2)(2x^2-x-1)(x^2-1) & = 0 \\ (x+2)(x-1)(2x+1)(x-1)(x+1)(x-1) & = 0 \\ (x+2)(2x+1)(x+1)(x-1)^3 & = 0 \\ \implies x & = -2, \ -\frac 12, \ - 1, \ 1 \end{aligned}$

$\implies -(a+b+c+d) = - \left(-2 -\dfrac 12 - 1 + 1\right) = \boxed{2.5}$

$(x^2 + x - 2)^3 + (2x^2 - x - 1)^3 = 27(x^2 - 1)^3$

$(x - 1)^3 (x + 2)^3 + (x - 1)^3 (2x + 1)^3 = 27 (x - 1)^3 (x + 1)^3$

$(x - 1)^3 ( 27(x + 1)^3 - (x + 2)^3 - (2x + 1)^3 ) = 0$

$(x - 1)^3 ( 18x^3 + 63x^2 + 63x + 18 ) = 0$ .... (I)

$18x^3 + 63x^2 + 63x + 18 =9(2x^3 + 7x^2 + 7x + 2) =$

$= 9(x+1)(2x^2 + 5x + 2) = 9(x+1)(2x + 1)(x + 2)$

$9(x - 1)^3 (x+1)(2x + 1)(x + 2) = 0$

Hence, the four distinct real roots do exist and they are:

a = 1 , b = -1 , c = -0.5 , d = -2 .

Therefore, our answer should be:

$-(a + b + c + d) = - ( 1 + (-1) + (-0.5) + (-2) ) = \boxed {2.5}$