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Question Number 19192 by Tinkutara last updated on 06/Aug/17

$$\mathrm{Find}\mathrm{a}\mathrm{natural}\mathrm{number}{}^{\prime }{\mathrm{n}}^{\prime }\mathrm{such}\mathrm{that}$$$$3^9+3^{12}+3^{15}+3^n\mathrm{is}\mathrm{a}\mathrm{perfect}\mathrm{cube}\mathrm{of}$$$$\mathrm{an}\mathrm{integer}.$$

Commented by Tinkutara last updated on 07/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Commented by RasheedSindhi last updated on 07/Aug/17

$$3^9+3^{12}+3^{15}+3^{\mathrm{n}}$$$$\mathrm{n}=14$$$$3^9+3^{12}+3^{15}+3^{14}$$$$=3^9+3^{12}+3^{14}+3^{15}$$$$={\left(3^3\right)}^3+3{\left(3^3\right)}^2\left(3^5\right)+3\left(3^3\right){\left(3^5\right)}^2+{\left(3^5\right)}^3$$$$={(3^3+3^5)}^3$$

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