Question

$\frac{ \sqrt[3]{512 } + \sqrt{216} }{ \sqrt[3]{343} }$

Answer 1

Hello!
Here is your answer

$512 = 2 {}^{9} \\ 216 = 6 {}^{3} \\ 343 = 7 {}^{3}$
From the above information the question turns out into

$\sqrt[3]{2 {}^{9} } + \sqrt{6 {}^{3} } \div \sqrt[3]{7 {}^{3} }$
We know that

√α = α^1/2

$2 {}^{9 \div 3} + 6 {}^{3 \div 2} \div 7 {}^{3 \div 3}$
$6 {}^{3} = 6 {}^{2} \times 6$
$2 {}^{3} + (6 {}^{2} \times 6) {}^{1 \div 2} \div 7$
$8 + 6 \sqrt{6} \div 7$