Question

How too do tjis sum pls sum body​

Answer 1

I assume you're referring to 25.

Find the prime factorizations of the numbers under the cube root:

`40=2^3\cdot5`

`625=5^4`

`320=2^6\cdot5`

The cube roots share a common factor of
`\sqrt[3]5`, so that

`2\sqrt[3]{40}+3\sqrt[3]{625}+4\sqrt[3]{320}=2\sqrt[3]{2^3\cdot5}+3\sqrt[3]{5^4}+4\sqrt[3]{2^6\cdot5}`

`=\sqrt[3]5\left(2\sqrt[3]{2^3}+3\sqrt[3]{5^3}+4\sqrt[3]{2^6}\right)`

We can simplify the cube roots now, since for any
`x` we have
`\sqrt[3]{x^3}=x`. So

`=\sqrt[3]5\left(2\cdot2+3\cdot5+4\cdot2^2\right)`

`=\sqrt[3]5\left(4+15+16\right)`

`=35\sqrt[3]5`