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Roots
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June 15, 2021 10:23 am
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(@armor)
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Given a=\sqrt[3]{4} +\sqrt[3]{2}+1,   find the value of  \tfrac{3}{a}+\tfrac{3}{a^2}+\tfrac{1}{a^3}.

 

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June 15, 2021 10:23 am
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(@armor)
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(\sqrt[3]{2}-1)a=(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)=2-1=1

\Rightarrow \tfrac{1}{a}=\sqrt[3]{2}-1,   

thus   \tfrac{3}{a}+\tfrac{3}{a^2}+\tfrac{1}{a^3}=\tfrac{3a^2+3a+1}{a^3}

=\tfrac{a^3+3a^2+3a+1-a^3}{a^3}

(\tfrac{a+1}{a})^3-1=(1+\tfrac{1}{a})^3-1=2-1=1.

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