if sin a+cos a=m then prove that sin⁶a +cos⁶a is equal to ( 4-3(m²-1)² )/4

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{sina}+\mathrm{cosa}=\mathrm{m} \\ \mathrm{Squarring} \mathrm{both} \mathrm{sides} \mathrm{we} \mathrm{get}: \\ {\left(\mathrm{sina}+\mathrm{cosa}\right)}^2={\mathrm{m}}^2 \\ \Rightarrow {\sin }^2\mathrm{a}+{\cos }^2\mathrm{a}+2\mathrm{sina}.\mathrm{cosa}={\mathrm{m}}^2 \\ \Rightarrow 1+2\mathrm{sina}.\mathrm{cosa}={\mathrm{m}}^2 \left[\mathrm{As} {\sin }^2\mathrm{a}+{\cos }^2\mathrm{a}=1\right] \\ \Rightarrow \mathrm{sina}.\mathrm{cosa}=\frac{{\mathrm{m}}^2-1}{2}.....\left(\mathrm{i}\right) \\ {\sin }^6\mathrm{a}+{\cos }^6\mathrm{a} \\ ={\left({\sin }^2\mathrm{a}\right)}^3+{\left({\cos }^2\mathrm{a}\right)}^3 \\ =\left({\sin }^2\mathrm{a}+{\cos }^2\mathrm{a}\right)\left({\sin }^4\mathrm{a}+{\cos }^4\mathrm{a}-{\sin }^2\mathrm{a} {\cos }^2\mathrm{a}\right) \\ =\left(1\right)\left\{{\left({\sin }^2\mathrm{a}\right)}^2+{\left({\cos }^2\mathrm{a}\right)}^2-{\sin }^2\mathrm{a} {\cos }^2\mathrm{a}\right\} \\ ={\left({\sin }^2\mathrm{a}\right)}^2+{\left({\cos }^2\mathrm{a}\right)}^2-{\sin }^2\mathrm{a} {\cos }^2\mathrm{a} \\ ={\left({\sin }^2\mathrm{a}+{\cos }^2\mathrm{a}\right)}^2-2{\sin }^2\mathrm{a} {\cos }^2\mathrm{a}-{\sin }^2\mathrm{a} {\cos }^2\mathrm{a} \left[\mathrm{As} {\mathrm{A}}^2+{\mathrm{B}}^2={\left(\mathrm{A}+\mathrm{B}\right)}^2-2\mathrm{AB}\right] \\ =1-3{\sin }^2\mathrm{a} {\cos }^2\mathrm{a} \\ =1-3{\left(\mathrm{sina}.\mathrm{cosa}\right)}^2 \\ =1-3{\left(\frac{{\mathrm{m}}^2-1}{2}\right)}^2 \left[\mathrm{By} \left(\mathrm{i}\right)\right] \\ =1-\frac{3{\left({\mathrm{m}}^2-1\right)}^2}{4} \\ \mathbf{\Rightarrow }{\mathbf{sin}}^{\mathbf{6}}\mathbf{a}\mathbf{+}{\mathbf{cos}}^{\mathbf{6}}\mathbf{a}\mathbf{=}\frac{\mathbf{4}\mathbf{-}\mathbf{3}{\left({\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}}{\mathbf{4}} \end{gathered}$$

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