KcetBinomial TheoremKCET 2026MCQ+1 / -02026The value at $x = 2$ for $\dfrac{x^3 + 3x^2 + 3x + 1}{x^4 + 4x^3 + 6x^2 +4x + 1}$A$3$B$\dfrac{25}{61}$C$\dfrac{1}{3}$D$\dfrac{19}{73}$Check AnswerClear SelectionReveal AnswerShow Explanation
