Q.The term independent of $x$ in the expansion of $\left( x + \dfrac{1}{x} \right)^{12} \left( x^2 + \dfrac{1}{x^2} \right)^8 \left( x^3 + \dfrac{1}{x^3} \right)^4$ is

\binom{12}{6} \binom{8}{4} \binom{4}{2}

\binom{12}{6} \binom{16}{8} \binom{12}{6}

\binom{36}{18}

\binom{24}{12}

The correct solution involves applying the fundamental concept to derive the final value step by step...

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Q.The term independent of xxx in the expansion of (x+1x)12(x2+1x2)8(x3+1x3)4\left( x + \dfrac{1}{x} \right)^{12} \left( x^2 + \dfrac{1}{x^2} \right)^8 \left( x^3 + \dfrac{1}{x^3} \right)^4(x+x1​)12(x2+x21​)8(x3+x31​)4 is