I assume you're referring to 25.
Find the prime factorizations of the numbers under the cube root:
[tex]40=2^3\cdot5[/tex]
[tex]625=5^4[/tex]
[tex]320=2^6\cdot5[/tex]
The cube roots share a common factor of [tex]\sqrt[3]5[/tex], so that
[tex]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}[/tex]
[tex]=\sqrt[3]5\left(2\sqrt[3]{2^3}+3\sqrt[3]{5^3}+4\sqrt[3]{2^6}\right)[/tex]
We can simplify the cube roots now, since for any [tex]x[/tex] we have [tex]\sqrt[3]{x^3}=x[/tex]. So
[tex]=\sqrt[3]5\left(2\cdot2+3\cdot5+4\cdot2^2\right)[/tex]
[tex]=\sqrt[3]5\left(4+15+16\right)[/tex]
[tex]=35\sqrt[3]5[/tex]
