8 < ∛a < 9
We need to solve for a.
Let's get the 1/3th root of the whole equation:
[tex] \sqrt[1/3]{8} \ \textless \ \sqrt[1/3]{cubic-root-of-x} \ \textless \ \sqrt[1/3]{9} [/tex]
[tex] \sqrt[1/3]{cubic-root-of-x} = x[/tex]
[tex] \sqrt[1/3]{8} = 512[/tex]
[tex] \sqrt[1/3]{9} = 712[/tex]
So
[tex] \sqrt[1/3]{8} \ \textless \ \sqrt[1/3]{cubic-root-of-x} \ \textless \ \sqrt[1/3]{9} [/tex]
512 < x < 729
Now from the values we got, 679 is between 512 and 729.
So 8 < ∛679 < 9 (∛679 ≈ 8.78) (Answer A)
Hope this helps! :D
