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

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

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

$=35\sqrt[3]5$