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If $u_{0}=8, u_{1}=3, u_{2}=12, u_{3}=51$, then the value of $\Delta^{3} u_{0}$ is
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Verified Answer
The correct answer is:
16
$\Delta^{3} u_{0}=(E-1)^{3} u_{0}$
$$
\begin{array}{l}
=\left(E^{3}-3 E^{2}+3 E-1\right) u_{0}=u_{3}-3 u_{2}+3 u_{1}-u_{0} \\
=51-3 \times 12+3 \times 3-8=16
\end{array}
$$
$$
\begin{array}{l}
=\left(E^{3}-3 E^{2}+3 E-1\right) u_{0}=u_{3}-3 u_{2}+3 u_{1}-u_{0} \\
=51-3 \times 12+3 \times 3-8=16
\end{array}
$$
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