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๐ฐ Now use $ p(1) $: $ 0 + 2 + 1 + d = 3 \Rightarrow d = 0 $.
๐ฐ Thus, $ p(x) = 2x^2 + x $. Then $ p(0) = 0 $.
๐ฐ But waitโthis is quadratic, but the problem states $ p(x) $ is cubic. Contradiction? Not necessarilyโ$ a = 0 $ is allowed if the cubic coefficient is zero, but the problem says cubic, implying degree exactly 3. So we must assume it's cubic, but our solution gives $ a = 0 $. That means the data fits a quadratic, but we are told it's cubic. So either the model is misclassified, or we must accept the interpolating polynomial, regardless of degree. Since the interpolation yields a unique cubic (degree at most 3), and we found $ a = 0 $, the polynomial is actually quadratic. But the problem says cubic, so we must reconsider.
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