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Question Number 186963 by 073 last updated on 12/Feb/23

Commented by 073 last updated on 12/Feb/23

$$\mathrm{solotion}\mathrm{please}??$$$$\mathrm{need}\mathrm{it}$$

Answered by Rasheed.Sindhi last updated on 13/Feb/23

$$\overline{)\begin{array}{cccccc}\ast & \mathrm{K}& \mathrm{E}& \mathrm{M}& \mathrm{A}& \mathrm{L}\\ \mathrm{K}& \mathrm{E}& \mathrm{M}& \mathrm{A}& \mathrm{L}& \mathrm{K}\\ \mathrm{E}& \mathrm{M}& \mathrm{A}& \mathrm{L}& \mathrm{K}& \mathrm{E}\\ \mathrm{M}& \mathrm{A}& \mathrm{L}& \mathrm{K}& \mathrm{E}& \mathrm{M}\\ \mathrm{A}& \mathrm{L}& \mathrm{K}& \mathrm{E}& \mathrm{M}& \mathrm{A}\\ \mathrm{L}& \mathrm{K}& \mathrm{E}& \mathrm{M}& \mathrm{A}& \mathrm{L}\end{array}}$$$${\mathrm{K}}^3\ast ({\mathrm{L}}^{-5}\ast \mathrm{E})=?$$$${\mathrm{K}}^3=(\mathrm{K}\ast \mathrm{K})\ast \mathrm{K}=\mathrm{E}\ast \mathrm{K}=\mathrm{M}$$$${\mathrm{L}}^{-5}={\left({\mathrm{L}}^{-1}\right)}^5={\left({\mathrm{L}}^5\right)}^{-1}$$$$\left({\mathrm{X}}^{-1}meansinverseof\mathrm{X}\right)$$$$Itcanbeobservedthat$$$$\mathrm{X}\ast \mathrm{L}=\mathrm{X},where\mathrm{X}\in \{\mathrm{K},\mathrm{E},\mathrm{M},\mathrm{A},\mathrm{L}\}$$$$\therefore \mathrm{L}isan\mathit{i}\mathit{d}\mathit{e}\mathit{n}\mathit{t}\mathit{i}\mathit{t}\mathit{y}\mathit{e}\mathit{l}\mathit{e}\mathit{m}\mathit{e}\mathit{n}\mathit{t}wrt\ast$$$$\because \mathrm{K}\ast \mathrm{A}=\mathrm{L}$$$$\therefore {\mathrm{K}}^{-1}=\mathrm{A}$$$$Similarly{\mathrm{L}}^{-1}=\mathrm{L}$$$${\mathrm{L}}^{-5}={\left({\mathrm{L}}^{-1}\right)}^5={\mathrm{L}}^5=\mathrm{L}$$$$$$$${\mathrm{K}}^3\ast ({\mathrm{L}}^{-5}\ast \mathrm{E})=\mathrm{M}\ast (\mathrm{L}\ast \mathrm{E})$$$$=\mathrm{M}\ast \mathrm{E}=\mathrm{L}$$

Commented by 073 last updated on 14/Feb/23

$$\mathrm{nice}\mathrm{solution}$$$$\mathrm{thanks}\mathrm{alot}$$

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