Question

simplify.
\\(\sqrt3{270} + 2\sqrt3{10}\\)
\\(?\sqrt3{\square}\\)

Explanation:

Step1: Factor 270

Factor 270 to find a perfect cube factor. We know that \(270 = 27\times10\), and \(27 = 3^3\). So, \(\sqrt[3]{270}=\sqrt[3]{27\times10}\).
Using the property of cube roots \(\sqrt[3]{ab}=\sqrt[3]{a}\times\sqrt[3]{b}\) (where \(a = 27\) and \(b = 10\)), we get \(\sqrt[3]{27\times10}=\sqrt[3]{27}\times\sqrt[3]{10}\).
Since \(\sqrt[3]{27}=3\) (because \(3^3 = 27\)), then \(\sqrt[3]{270}=3\sqrt[3]{10}\).

Step2: Substitute and combine like terms

Now substitute \(\sqrt[3]{270}=3\sqrt[3]{10}\) into the original expression \(\sqrt[3]{270}+2\sqrt[3]{10}\).
We get \(3\sqrt[3]{10}+2\sqrt[3]{10}\).
Since these are like terms (both have \(\sqrt[3]{10}\)), we can combine the coefficients: \((3 + 2)\sqrt[3]{10}=5\sqrt[3]{10}\).

Answer:

\(5\sqrt[3]{10}\) (So the first box is \(5\) and the second box is \(10\))