Question
Evaluate the following: $\sqrt[3]{\text{a}24}+\sqrt[3]{0.008}+\sqrt[3]{0.0064}$
✓
Answer
To evaluate the value of the given expression, we need to proceed as follows: $\sqrt[3]{\text{a}24}+\sqrt[3]{0.008}+\sqrt[3]{0.0064}$
$=\sqrt[3]{3\times3\times3}+\sqrt[3]{\frac{8}{1000}}+\sqrt[3]{\frac{64}{1000}}$
$=\sqrt[3]{3\times3\times3}+{\frac{\sqrt[3]{8}}{\sqrt[3]{1000}}}+{\frac{\sqrt[3]{64}}{\sqrt[3]{1000}}}$
$=\sqrt[3]{3\times3\times3}+{\frac{\sqrt[3]{2\times2\times2}}{\sqrt[3]{1000}}}+{\frac{\sqrt[3]{4\times4\times4}}{\sqrt[3]{1000}}}$
$=3+\frac{2}{10}+\frac{4}{10}$
$=3+0.2+0.4$
$=3.6$ Thus, the answer is $3.6.$
$=\sqrt[3]{3\times3\times3}+\sqrt[3]{\frac{8}{1000}}+\sqrt[3]{\frac{64}{1000}}$
$=\sqrt[3]{3\times3\times3}+{\frac{\sqrt[3]{8}}{\sqrt[3]{1000}}}+{\frac{\sqrt[3]{64}}{\sqrt[3]{1000}}}$
$=\sqrt[3]{3\times3\times3}+{\frac{\sqrt[3]{2\times2\times2}}{\sqrt[3]{1000}}}+{\frac{\sqrt[3]{4\times4\times4}}{\sqrt[3]{1000}}}$
$=3+\frac{2}{10}+\frac{4}{10}$
$=3+0.2+0.4$
$=3.6$ Thus, the answer is $3.6.$
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