Question

$$( \mu_{\bar{x}} \times \alpha )+( \sigma_{\bar{x}} \sigma + \alpha \alpha ) \times (4 \mu +51)$$

Answer

$$m*a^2*l*p*h*u_bar(x)+(4*m*u+51)*(-s^2*g^2*m^2*a*a_bar(x)+aa*a^2*l^2*p^2*h^2)$$

Solution


Remove parentheses.
\[mu_bar(x)alpha+(s\imath gma_bar(x)s\imath gma+alphaalpha)(4mu+51)\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[mu_bar(x)alpha+({s}^{2}{\imath }^{2}{g}^{2}{m}^{2}a_bar(x)a+alphaalpha)(4mu+51)\]
Use Square Rule: \({i}^{2}=-1\).
\[mu_bar(x)alpha+({s}^{2}\times -1\times {g}^{2}{m}^{2}a_bar(x)a+alphaalpha)(4mu+51)\]
Simplify  \({s}^{2}\times -1\times {g}^{2}{m}^{2}a_bar(x)a\)  to  \({s}^{2}\times -{g}^{2}{m}^{2}a_bar(x)a\).
\[mu_bar(x)alpha+({s}^{2}\times -{g}^{2}{m}^{2}a_bar(x)a+alphaalpha)(4mu+51)\]
Regroup terms.
\[mu_bar(x)alpha+(-{s}^{2}{g}^{2}{m}^{2}aa_bar(x)+alphaalpha)(4mu+51)\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[mu_bar(x)alpha+(-{s}^{2}{g}^{2}{m}^{2}aa_bar(x)+{a}^{2}{l}^{2}{p}^{2}{h}^{2}aa)(4mu+51)\]
Regroup terms.
\[mu_bar(x)alpha+(-{s}^{2}{g}^{2}{m}^{2}aa_bar(x)+aa{a}^{2}{l}^{2}{p}^{2}{h}^{2})(4mu+51)\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[mu_bar(x){a}^{2}lph+(-{s}^{2}{g}^{2}{m}^{2}aa_bar(x)+aa{a}^{2}{l}^{2}{p}^{2}{h}^{2})(4mu+51)\]
Regroup terms.
\[m{a}^{2}lphu_bar(x)+(-{s}^{2}{g}^{2}{m}^{2}aa_bar(x)+aa{a}^{2}{l}^{2}{p}^{2}{h}^{2})(4mu+51)\]
Regroup terms.
\[m{a}^{2}lphu_bar(x)+(4mu+51)(-{s}^{2}{g}^{2}{m}^{2}aa_bar(x)+aa{a}^{2}{l}^{2}{p}^{2}{h}^{2})\]