Will you cube each term?

Let $a=x^{2}+x-2=(x+2)(x-1)$ and $b=2x^{2}-x-1=(2x+1)(x-1)$

$a^{3}+b^{3}=(a+b)^{3} \\ab(a+b)=0 \\(x+2)(x-1)(2x+1)(x-1)(x^{2}-1)=0\\ x=1,-1,-2,\frac{-1}{2}$

According to the question answer is $\boxed{1-1-2-0.5=-2.5}$

$\begin{aligned} (x^2+x-2)^3+(2x^2-x-1)^3 & = 27(x^2-1)^3 \\ (x-1)^3(x+2)^3 + (x-1)^3(2x+1)^3 & = 27(x-1)^3 (x+1)^3 \end{aligned}$

We note that $x = 1$ is a root and that:

$\begin{aligned} (\color{#3D99F6}{x+2})^3 + (\color{#D61F06}{2x+1})^3 & = 27(x+1)^3 = (3x+3)^3 = (\color{#3D99F6}{x+2} +\color{#D61F06}{2x+1})^3 \\ \Rightarrow \color{#3D99F6}{a}^3 + \color{#D61F06}{b}^3 & = (\color{#3D99F6}{a} + \color{#D61F06}{b})^3 \\ \Rightarrow ab(a+b) & = 0 \\ (x+2)(2x+1)(3x+3) & = 0 \\ \Rightarrow (x+2)(2x+1)(x+1) & = 0 \end{aligned}$

This implies that all the four real roots are $1$ , $-1$ , $-2$ and $-\frac{1}{2}$ and their sum is $\boxed{-2.5}$ .

Note first that if we were to expand all the terms we would get a degree $6$ polynomial, so we can expect $6$ solutions, although we cannot tell at this point if all of them will be real.

Now the equation can be rewritten as $((x - 1)(x + 2))^{3} + ((2x + 1)(x - 1))^{3} = 3^{3}((x - 1)(x + 1))^{3}.$

Thus $x = 1$ is a root of multiplicity $3,$ and so we are looking for $3$ more roots, (not necessarily all real). Dividing through by $(x - 1)^{3}$ gives us the equation

$(x + 2)^{3} + (2x + 1)^{3} = (3x + 3)^{3}.$

Now we could play a hunch and suppose that $(x + 2), (2x + 1)$ and $(3x + 3)$ are all integers, in which case, by Fermat's Last Theorem, not all of them can be positive integers, (or all negative integers, for that matter). So if we were to successively set each of these terms equal to $0,$ we quickly notice that $x = -2, x = -\frac{1}{2}$ and $x = -1$ all satisfy this last equation. This indicates that we have found all the real roots, the distinct roots summing to $1 + (-1) + (-2) + (-\frac{1}{2}) = -\frac{5}{2} = \boxed{-2.5}.$

Less intuitively, we can expand the last equation to end up with

$x^{3} + 6x^{2} + 12x + 8 + 8x^{3} + 12x^{2} + 6x + 1 = 27x^{3} + 81x^{2} + 81x + 27$

$\Longrightarrow 18x^{3} + 63x^{2} + 63x + 18 = 0 \Longrightarrow x^{3} + \dfrac{7}{2}x^{2} + \dfrac{7}{2}x + 1 = 0.$

Now by Vieta's we know that the sum of the remaining roots is $-\dfrac{7}{2},$ but we don't know yet if all the remaining roots are real. By Descartes' Rule of Signs we know that none of the remaining roots are positive and that either $3$ or $1$ are negative. However, by the symmetry of this last equation we see that $x = -1$ is a solution, allowing us to further factor to

$(x + 1)(x^{2} + \frac{5}{2}x + 1) = (x + 1)(x + \frac{1}{2})(x + 2) = 0,$

which confirms that the remaining roots are all negative reals which sum to $-\dfrac{7}{2}.$ Adding the root $x = 1$ to this gives the same answer as before, namely $1 - \dfrac{7}{2} = -\dfrac{5}{2} = \boxed{-2.5}.$

It's easy to observe that if each of the expression which is cubed is taken and added, then the result is 0. We know that when a+b+c=0, then a^3 + b^3 + c^3= 3abc.

Also, here a^3 + b^3 + c^3 = 0. So, 3abc=0. This enables us to factor the given terns and put the whole thing = 0. So we have distinct solutions of 1,-1,-2,-0.5 .So the sum is -2.5

$(x^{2}+x-2)^{3}+(2x^{2}-x-1)^{3}+(3-3x^{2})^{3}=0$ As $(x^{2}+x-2)+(2x^{2}-x-1)+(3-3x^{2})=0$ we have that the inital ecuation becomes $9(x^{2}+x-2)(2x^{2}-x-1)(1-x^{2})=0$ where the distinct solutions are $1;-1;-2;-\frac{1}{2}$

fermat's last theorem states that "there is no non trivial solution to the equation a^n+b^n=c^n for n>=3",hence for the above equation we have a=0,b=0,c=0,therefore on solving we get x=1,-1,-2,-0.5 =>-2.5

Upon factorising the above term we get roots which are $1, -\frac {1}{2},-2,-1and 1$ out of which we have to find the sum of $1 , -\frac {1}{2},-2and-1$ which is -2•5 as we have to take distinct roots we have neglected 1,1

Let a= $x^{2}+x-2$ b= $2x^{2}-x-1$ c= $-3 (x^{2}-1)$ Then a+b+c= $x^{2} +x-2 +(2x^{2}-x-1)+(-3x^{2}-3)$ = 0

Hence $a^{3}+b^{3}+c^{3} -3abc$ = (a+b+c) $a^{2} +b^{2}+ c^{2}-ab-bc-ca$ =0

Hence $a^{3} +b^{3}+c^{3}$ = 3abc = 0

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

Hence x= 1-1-2-0.5 = -2.5