Question

$$f(3)^{\prime}=\binom{4}{3}(\frac{1}{4})^{3}(\frac{3}{4})^{4-3}=$$

Answer

$$f=-3^(1-prime)+(117992*34^4*IM*b*n*o*m)/3^prime$$

Solution


Simplify  \({14}^{3}\)  to  \(2744\).
\[f\times {3}^{prime}=b\imath nom\times 43\times 2744\times {34}^{4}-3\]
Regroup terms.
\[{3}^{prime}f=b\imath nom\times 43\times 2744\times {34}^{4}-3\]
Simplify  \(b\imath nom\times 43\times 2744\times {34}^{4}\)  to  \(117992bnom\imath \times {34}^{4}\).
\[{3}^{prime}f=117992bnom\imath \times {34}^{4}-3\]
Regroup terms.
\[{3}^{prime}f=117992\times {34}^{4}\imath bnom-3\]
Regroup terms.
\[{3}^{prime}f=-3+117992\times {34}^{4}\imath bnom\]
Divide both sides by \({3}^{prime}\).
\[f=\frac{-3+117992\times {34}^{4}\imath bnom}{{3}^{prime}}\]
Simplify  \(\frac{-3+117992\times {34}^{4}\imath bnom}{{3}^{prime}}\)  to  \(-{3}^{1-prime}+\frac{117992\times {34}^{4}\imath bnom}{{3}^{prime}}\).
\[f=-{3}^{1-prime}+\frac{117992\times {34}^{4}\imath bnom}{{3}^{prime}}\]