Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{{x}^{3}}{3}=\frac{{x}^{2}}{2}+x+\frac{{x}^{8}{x}^{2}}{2\times 3}+\frac{x}{4}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[\frac{{x}^{3}}{3}=\frac{{x}^{2}}{2}+x+\frac{{x}^{8+2}}{2\times 3}+\frac{x}{4}\]
Simplify \(8+2\) to \(10\).
\[\frac{{x}^{3}}{3}=\frac{{x}^{2}}{2}+x+\frac{{x}^{10}}{2\times 3}+\frac{x}{4}\]
Simplify \(2\times 3\) to \(6\).
\[\frac{{x}^{3}}{3}=\frac{{x}^{2}}{2}+x+\frac{{x}^{10}}{6}+\frac{x}{4}\]
Simplify \(\frac{{x}^{2}}{2}+x+\frac{{x}^{10}}{6}+\frac{x}{4}\) to \(\frac{{x}^{2}}{2}+\frac{5x}{4}+\frac{{x}^{10}}{6}\).
\[\frac{{x}^{3}}{3}=\frac{{x}^{2}}{2}+\frac{5x}{4}+\frac{{x}^{10}}{6}\]
Multiply both sides by \(12\) (the LCM of \(3, 2, 4, 6\)).
\[4{x}^{3}=6{x}^{2}+15x+2{x}^{10}\]
Move all terms to one side.
\[4{x}^{3}-6{x}^{2}-15x-2{x}^{10}=0\]
Factor out the common term \(x\).
\[x(4{x}^{2}-6x-15-2{x}^{9})=0\]
No root was found algebraically. However, the following root(s) were found by numerical methods.
\[x=-1.076981,0\]
x=-1.0769813537598,0
