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

Find a natural number ′n′ such that  3^9  + 3^(12)  + 3^(15)  + 3^n  is a perfect cube of  an integer.

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

Commented by Tinkutara last updated on 07/Aug/17

Thank you very much Sir!

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

Commented by RasheedSindhi last updated on 07/Aug/17

3^9 +3^(12) +3^(15) +3^n   n=14  3^9 +3^(12) +3^(15) +3^(14)   =3^9 +3^(12) +3^(14) +3^(15)   =(3^3 )^3 +3(3^3 )^2 (3^5 )+3(3^3 )(3^5 )^2 +(3^5 )^3   =(3^3 +3^5 )^3

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

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