if x^2+1/2x=cos theta then x^6+1/2x^3=

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$$\begin{gathered} {\mathrm{x}}^2+\frac{1}{2\mathrm{x}}=\mathrm{cos\theta }...\left(\mathrm{ii}\right) \\ \mathrm{Cube} \mathrm{both} \mathrm{sides} \\ {\left({\mathrm{x}}^2+\frac{1}{2\mathrm{x}}\right)}^3={\cos }^3\mathrm{\theta } \\ \mathrm{As} {\left(\mathrm{a}+\mathrm{b}\right)}^3={\mathrm{a}}^3+{\mathrm{b}}^3+3{\mathrm{a}}^2\mathrm{b}+3{\mathrm{ab}}^2 \\ \Rightarrow {\mathrm{x}}^6+\frac{1}{8{\mathrm{x}}^3}+3{\mathrm{x}}^4.\left(\frac{1}{2\mathrm{x}}\right)+3{\mathrm{x}}^2.\frac{1}{4{\mathrm{x}}^2}={\cos }^3\mathrm{\theta } \\ \Rightarrow {\mathrm{x}}^6+\frac{1}{8{\mathrm{x}}^3}+\frac{3{\mathrm{x}}^3}{2}+\frac{3}{4}={\cos }^3\mathrm{\theta } \\ \mathrm{Multiply} \mathrm{both} \mathrm{sides} 4 \\ \Rightarrow 4{\mathrm{x}}^6+\frac{1}{2{\mathrm{x}}^3}+6{\mathrm{x}}^3+3=4{\cos }^3\mathrm{\theta }....\left(\mathrm{ii}\right) \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \cos 3\mathrm{\theta }=4{\cos }^3\mathrm{\theta }-3\mathrm{cos\theta } \\ =4{\mathrm{x}}^6+\frac{1}{2{\mathrm{x}}^3}+6{\mathrm{x}}^3+3-3\left({\mathrm{x}}^2+\frac{1}{2\mathrm{x}}\right) \left[\mathrm{Use} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right)\right] \\ =4{\mathrm{x}}^6+\frac{1}{2{\mathrm{x}}^3}+6{\mathrm{x}}^3-3{\mathrm{x}}^2+3-\frac{3}{2\mathrm{x}} \\ ={\mathrm{x}}^6+\frac{1}{2{\mathrm{x}}^3}+3{\mathrm{x}}^6+6{\mathrm{x}}^3-3{\mathrm{x}}^2+3-\frac{3}{2\mathrm{x}} \\ \Rightarrow \cos 3\mathrm{\theta }=\left({\mathrm{x}}^6+\frac{1}{2{\mathrm{x}}^3}\right)+3\left({\mathrm{x}}^6+2{\mathrm{x}}^3-{\mathrm{x}}^2+1-\frac{1}{2\mathrm{x}}\right) \\ \Rightarrow \left({\mathrm{x}}^6+\frac{1}{2{\mathrm{x}}^3}\right)=\cos 3\mathrm{\theta }-3\left({\mathrm{x}}^6+2{\mathrm{x}}^3-{\mathrm{x}}^2+1-\frac{1}{2\mathrm{x}}\right) \\ =\cos 3\mathrm{\theta }-3\left\{{\mathrm{x}}^6+2{\mathrm{x}}^3+1-\left({\mathrm{x}}^2+\frac{1}{2\mathrm{x}}\right)\right\} \\ =\cos 3\mathrm{\theta }-3\left\{{\mathrm{x}}^6+2{\mathrm{x}}^3+1-\mathrm{cos\theta }\right\} \\ =\cos 3\mathrm{\theta }+3\mathrm{cos\theta }-3\left({\mathrm{x}}^6+2{\mathrm{x}}^3+1\right) \\ \left({\mathrm{x}}^6+\frac{1}{2{\mathrm{x}}^3}\right)=\cos 3\mathrm{\theta }+3\mathrm{cos\theta }-3{\left({\mathrm{x}}^3+1\right)}^2 \end{gathered}$$

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