Question

$$\left. \begin{array} { l } { \frac { 1 } { n } \frac { 1 } { b } \frac { 1 } { b } \frac { 1 } { 12 } } \\ { m a r i c } \\ { 6 } \\ { n } \\ { 0 f \circ b } \end{array} \right.$$

Answer

$$1568*Fi*b*n*d*a*r*t*h+cPr*e*m*a*t*o^3*g*r*s^2*n*f$$

Solution


Simplify  \(\frac{14}{1}\)  to  \(14\).
\[14b\times 112Findarth-mat\imath cProgress\imath onof\]
Simplify  \(14b\times 112Findarth\)  to  \(1568bndarthFi\).
\[1568bndarthFi-mat\imath cProgress\imath onof\]
Regroup terms.
\[1568Fibndarth-mat\imath cProgress\imath onof\]
Regroup terms.
\[1568Fibndarth-matooogrssnf\imath cPre\imath \]
Simplify  \(matooogrssnf\imath cPre\imath \)  to  \(mat{o}^{3}gr{s}^{2}nf\imath cPre\imath \).
\[1568Fibndarth-mat{o}^{3}gr{s}^{2}nf\imath cPre\imath \]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[1568Fibndarth-mat{o}^{3}gr{s}^{2}nf{\imath }^{2}cPre\]
Use Square Rule: \({i}^{2}=-1\).
\[1568Fibndarth-mat{o}^{3}gr{s}^{2}nf\times -1\times cPre\]
Simplify  \(mat{o}^{3}gr{s}^{2}nf\times -1\times cPre\)  to  \(mat{o}^{3}gr{s}^{2}nf\times -cPre\).
\[1568Fibndarth-mat{o}^{3}gr{s}^{2}nf\times -cPre\]
Regroup terms.
\[1568Fibndarth-(-cPremat{o}^{3}gr{s}^{2}nf)\]
Remove parentheses.
\[1568Fibndarth+cPremat{o}^{3}gr{s}^{2}nf\]