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Sagot :
[tex]\sqrt[3]{343}+\frac{3}{4}\sqrt[3]{-8}\text{ =}\frac{11}{2}[/tex]
Explanation
The cube root of a number is the factor that we multiply by itself three times to get that number
so
Step 1
[tex]\sqrt[3]{343}+\frac{3}{4}\sqrt[3]{-8}[/tex][tex]\begin{gathered} \sqrt[3]{343}\text{ =7} \\ because,\cdot7\cdot7\cdot7=343 \\ \end{gathered}[/tex]and
[tex]\begin{gathered} \sqrt[3]{-8}\text{ =-2} \\ \text{because -2}\cdot-2\cdot-2=-8 \end{gathered}[/tex]Step 2
replace
[tex]\begin{gathered} \sqrt[3]{343}+\frac{3}{4}\sqrt[3]{-8} \\ 7+\frac{3}{4}(-2)=7-\frac{6}{4}=\frac{28-6}{4}=\frac{22}{4}=\frac{11}{2} \end{gathered}[/tex]so,the answer is
[tex]\sqrt[3]{343}+\frac{3}{4}\sqrt[3]{-8}\text{ =}\frac{11}{2}[/tex]I hope this helps you
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