Use Difference of Cubes: \({a}^{3}-{b}^{3}=(a-b)({a}^{2}+ab+{b}^{2})\).
\[l\imath mx->a\times \frac{(x-a)({x}^{2}+(x)(a)+{a}^{2})}{x-a}\]
Remove parentheses.
\[l\imath mx->a\times \frac{(x-a)({x}^{2}+xa+{a}^{2})}{x-a}\]
Cancel \(x-a\).
\[l\imath mx->a({x}^{2}+xa+{a}^{2})\]
Regroup terms.
\[-+l\imath mx>a({x}^{2}+xa+{a}^{2})\]
Divide both sides by \(\imath \).
\[-+lmx>\frac{a({x}^{2}+xa+{a}^{2})}{\imath }\]
Rationalize the denominator: \(\frac{a({x}^{2}+xa+{a}^{2})}{\imath } \cdot \frac{\imath }{\imath }=-a({x}^{2}+xa+{a}^{2})\imath \).
\[-+lmx>-a({x}^{2}+xa+{a}^{2})\imath \]
Regroup terms.
\[-+lmx>-\imath a({x}^{2}+xa+{a}^{2})\]
Divide both sides by \(m\).
\[-+lx>-\frac{\imath a({x}^{2}+xa+{a}^{2})}{m}\]
Divide both sides by \(x\).
\[-+l>-\frac{\frac{\imath a({x}^{2}+xa+{a}^{2})}{m}}{x}\]
Simplify \(\frac{\frac{\imath a({x}^{2}+xa+{a}^{2})}{m}}{x}\) to \(\frac{\imath a({x}^{2}+xa+{a}^{2})}{mx}\).
\[-+l>-\frac{\imath a({x}^{2}+xa+{a}^{2})}{mx}\]
Multiply both sides by \(-1\).
\[l<\frac{\imath a({x}^{2}+xa+{a}^{2})}{mx}\]
l<(IM*a*(x^2+x*a+a^2))/(m*x)
