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Consider

[tex] \rm\[ p(x)=x^{3}+3 x^{2}+3 x+1 \] \\ [/tex]

[tex] \text{ \( \therefore \quad \) By Remainder Theorem, the remainder is \: \( \tt p\left(\dfrac{1}{2}\right) \)}[/tex]

[tex] \[ \begin{aligned} \tt \therefore \quad p\left(\frac{1}{2}\right) & \tt=\left(\frac{1}{2}\right)^{3}+3\left(\frac{1}{2}\right)^{2}+3\left(\frac{1}{2}\right)+1 \\ \\ & \mathfrak{=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1=\frac{1+6+12+8}{8}=\frac{27}{8}} \\ \\ \color{magenta} \mathfrak{ \Rightarrow \quad p\left(\frac{1}{2}\right)} &\color{magenta} \mathfrak{=\frac{27}{8}} \end{aligned} \][/tex]

[tex] \\ \text{ હા છે ને...} \\ \text{કેમ છો.} \\\text{મજામાં }[/tex]