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If the sum of two particular roots of the equation $x^4-4 x^3$ $-7 x^2+22 x+24=0$ is equal to the sum of the remaining two roots, then the sum of the cubes of all the roots of this equation is
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$82$
$x^4-4 x^3-7 x^2+22 x+24=0$
$\begin{aligned} & \Rightarrow \quad(x+1)\left(x^3-5 x^2-2 x+24\right)=0 \\ & \Rightarrow \quad(x+1)(x+2)\left(x^2-7 x+12\right)=0 \\ & \Rightarrow \quad(x+1)(x+2)(x-3)(x-4)=0 \\ & \Rightarrow \quad x=-1,-2,3,4 \\ & \therefore \quad 3-1=4-2 \\ & \therefore \quad \alpha^3+\beta^3+\gamma^3+\delta^3=(3)^3+(4)^3+(-1)^3+(-2)^3=82 .\end{aligned}$
$\begin{aligned} & \Rightarrow \quad(x+1)\left(x^3-5 x^2-2 x+24\right)=0 \\ & \Rightarrow \quad(x+1)(x+2)\left(x^2-7 x+12\right)=0 \\ & \Rightarrow \quad(x+1)(x+2)(x-3)(x-4)=0 \\ & \Rightarrow \quad x=-1,-2,3,4 \\ & \therefore \quad 3-1=4-2 \\ & \therefore \quad \alpha^3+\beta^3+\gamma^3+\delta^3=(3)^3+(4)^3+(-1)^3+(-2)^3=82 .\end{aligned}$
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