sin⁴C/a + cos⁴C/b = 1/a+b is given , now  
prove that - sin⁸C/a³ + cos⁸C/b³ = 1/(a+b)³

Dear Student,
$$\begin{gathered} \frac{{\sin }^4 C}{a}+\frac{{\cos }^{4}C}{b}=\frac{1}{a+b} \\ Solution: We know that {\cos }^{4}x = {\left({\cos }^{2}x\right)}^2 = {\left(1 – {\sin }^{2}x\right)}^2 = 1 + {\sin }^{4}x – 2{\sin }^{2}x \\ So, \frac{{\sin }^4 C}{a}+ \frac{{\cos }^{4}C}{b} =\frac{{\displaystyle 1}}{\left(a+b\right)} \\ \Rightarrow \frac{{\sin }^4 C}{a} + \frac{{\displaystyle 1 + {\sin }^{4}C – 2{\sin }^{2}C}}{b} =\frac{{\displaystyle 1}}{\left(a+b\right)} \\ \Rightarrow \frac{b{\sin }^{4}C+a+a{\sin }^{4}C-2a{\sin }^{2}C}{ab}=\frac{{\displaystyle 1}}{\left(a+b\right)} \\ \Rightarrow \frac{{\displaystyle \left(a+b\right){\sin }^{4}C+a-2a{\sin }^{2}C}}{{\displaystyle ab}}=\frac{{\displaystyle 1}}{\left(a+b\right)} \\ \Rightarrow \left(a+b\right)\left(\left(a+b\right){\sin }^{4}C+a-2a{\sin }^{2}C\right)=ab \\ \Rightarrow {\left(a+b\right)}^2{\sin }^{4}C+a\left(a+b\right)-2a\left(a+b\right){\sin }^{2}C=ab \\ \Rightarrow {\left(a+b\right)}^2{\sin }^{4}C+a^2+ab-2a\left(a+b\right){\sin }^{2}C-ab=0 \left[U\sin g: {\left(a-b\right)}^2 = a^2+b^2-2ab\right] \\ \Rightarrow {\left(\left(a+b\right){\sin }^{2}C-a\right)}^2=0 \\ \Rightarrow \left(a+b\right){\sin }^{2}C-a = 0 \\ \Rightarrow {\sin }^{2}C=\frac{a}{a+b}...........\left(i\right) \\ \Rightarrow So, {\sin }^{8}C=\frac{{\displaystyle a^4}}{{\displaystyle {\left(a+b\right)}^4}} \\ \Rightarrow \frac{{\displaystyle {\sin }^{8}C}}{a^3}=\frac{{\displaystyle a}}{{\displaystyle {\left(a+b\right)}^4}}............\left(ii\right) \\ Now , {\cos }^{2}C=1-{\sin }^{2}C \left[{\sin }^{2}a+{\cos }^{2}a=1\right] \\ {\cos }^{2}C=1-\frac{a}{a+b} \left[from equation \left(i\right)\right] \\ {\cos }^{2}C=\frac{a+b-a}{a+b} \\ {\cos }^{2}C=\frac{b}{a+b} \\ \frac{{\displaystyle {\cos }^{8}C}}{{{\displaystyle b}}^3}=\frac{{\displaystyle b}}{{\displaystyle {\left(a+b\right)}^4}}............\left(iii\right) \\ Adding equation \left(ii\right) and \left(iii\right) \\ \Rightarrow \frac{{\displaystyle {\sin }^{8}C}}{a^3}+\frac{{\displaystyle {\cos }^{8}C}}{{\displaystyle b^3}}=\frac{a}{{\displaystyle {\left(a+b\right)}^4}}+\frac{{\displaystyle b}}{{\displaystyle {\left(a+b\right)}^4}} \\ \Rightarrow \frac{{\displaystyle {\sin }^{8}C}}{a^3}+\frac{{\displaystyle {\cos }^{8}C}}{{\displaystyle b^3}}=\frac{{\displaystyle a+b}}{{\displaystyle {\left(a+b\right)}^4}} \\ \Rightarrow \boxed{\frac{{\displaystyle {\sin }^{8}C}}{{\mathbf{a}}^{\mathbf{3}}}\mathbf{+}\frac{{\displaystyle {\cos }^{8}C}}{{\displaystyle b^3}}\mathbf{=}\frac{{\displaystyle 1}}{{\displaystyle {\left(a+b\right)}^3}}} \\ Hence Proved \end{gathered}$$

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