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B
−1+x+x2−3x3
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C
3x3−x2+x−1
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D
3x3+x2−x−1
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Solution
The correct option is C3x3−x2+x−1 Let, the term to be added be y. ⇒1–x+x2–2x3+y=x3 ⇒y=x3–(1–x+x2–2x3) =x3–1+x–x2+2x3 =3x3–x2+x–1.
Hence, y = 3x3–x2+x–1.