prove that

Question:

$\sqrt[3]{8+27}=\sqrt[3]{8}+\sqrt[3]{27}$

Solution:

False

We have, $\sqrt[3]{8+27} \neq \sqrt[3]{8}+\sqrt[3]{27}$

$\because$ $\sqrt[3]{8}+\sqrt[3]{27}=\sqrt[3]{2 \times 2 \times 2}+\sqrt[3]{3 \times 3 \times 3}=2+3=5$

But $\sqrt[3]{8+27}=\sqrt[3]{35} \neq \sqrt[3]{125}=5$

 

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