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(B) The given equation is $x^{4}-3x^{3}-2x^{2}+3x+1=10$,which simplifies to $x^{4}-3x^{3}-2x^{2}+3x-9=0$.Dividing by $x^{2}$ (since $x=0$ is not a roo

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(B) The given equation is $x^{4}-3x^{3}-2x^{2}+3x+1=10$,which simplifies to $x^{4}-3x^{3}-2x^{2}+3x-9=0$.Dividing by $x^{2}$ (since $x=0$ is not a root),we get $x^{2}-3x-2+\frac{3}{x}-\frac{9}{x^{2}}=0$.Rearranging terms: $(x^{2}-\frac{9}{x^{2}})-3(x-\frac{1}{x})-2=0$. This approach is complex,let's factorize directly.$x^{4}-3x^{3}-2x^{2}+3x-9=0$$x^{3}(x-3)-2(x^{2}-1.5x+4.5)$ is not direct. Let's group: $(x^{4}-9)-3x(x^{2}-1)-2x^{2}=0$.Actually,$x^{4}-3x^{3}-2x^{2}+3x-9=0$ can be written as $(x^{2}-3)(x^{2}-3x+3)=0$ is incorrect. Let's test roots: $x=3$ gives $81-81-18+9-9 
eq 0$. $x=-1$ gives $1+3-2-3-9 
eq 0$. $x=3$ is a root of $x^{2}-3x-3=0$?Let's use $x^{2}-3x-3=0$ and $x^{2}+3=0$. Roots are $x^{2}=3x+3$ and $x^{2}=-3$.For $x^{2}=-3$,$x = \pm i\sqrt{3}$. Cubes are $-3i\sqrt{3}$ and $3i\sqrt{3}$,sum $= 0$.For $x^{2}-3x-3=0$,roots $\alpha, \beta$ satisfy $\alpha+\beta=3, \alpha\beta=-3$.Sum of cubes $\alpha^{3}+\beta^{3} = (\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta) = 3^{3}-3(-3)(3) = 27+27 = 54$.Wait,the equation is $x^{4}-3x^{3}-2x^{2}+3x-9=0$. $(x^{2}-3)(x^{2}-3x+3)=0$ is $x^{4}-3x^{3}+3x^{2}-3x^{2}+9x-9 = x^{4}-3x^{3}+9x-9$. Not matching.Correct factorization: $(x^{2}+3)(x^{2}-3x-3)=0$.Roots of $x^{2}+3=0$ are $i\sqrt{3}, -i\sqrt{3}$. Cubes are $-3i\sqrt{3}, 3i\sqrt{3}$,sum $= 0$.Roots of $x^{2}-3x-3=0$ are $\alpha, \beta$. $\alpha+\beta=3, \alpha\beta=-3$.Sum of cubes $= (\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta) = 27-3(-3)(3) = 27+27 = 54$.

The sum of the cubes of all the roots of the equation $x^{4}-3x^{3}-2x^{2}+3x+1=10$ is

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