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Question Number 134763 by bramlexs22 last updated on 07/Mar/21

Given a= (3)^(1/3)  + (1/( (√3)))  Find the value of 3a^3 −9a+1.

$$\mathrm{Given}\:{a}=\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{3}{a}^{\mathrm{3}} −\mathrm{9}{a}+\mathrm{1}. \\ $$

Answered by EDWIN88 last updated on 07/Mar/21

= ((3(√2) (9)^(1/3) )/3) +((2(√3) ((243))^(1/6) )/9)−((26 (√3))/9) +((2 (3)^(1/3)  ((243))^(1/6) )/3)−((26 (3)^(1/3) )/3)+4

$$=\:\frac{\mathrm{3}\sqrt{\mathrm{2}}\:\sqrt[{\mathrm{3}}]{\mathrm{9}}}{\mathrm{3}}\:+\frac{\mathrm{2}\sqrt{\mathrm{3}}\:\sqrt[{\mathrm{6}}]{\mathrm{243}}}{\mathrm{9}}−\frac{\mathrm{26}\:\sqrt{\mathrm{3}}}{\mathrm{9}}\:+\frac{\mathrm{2}\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:\sqrt[{\mathrm{6}}]{\mathrm{243}}}{\mathrm{3}}−\frac{\mathrm{26}\:\sqrt[{\mathrm{3}}]{\mathrm{3}}}{\mathrm{3}}+\mathrm{4} \\ $$

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