Determine whether the number is divisible by 3. Give a reason based on the tests for divisibility
1 answer:
You might be interested in
So we have a 6 at the bottom, and the root is 3, so hmm how to take it out, simple enough, just let's get something to make the 6 a 6³, so it comes out of the root
so
![\bf \cfrac{2+\sqrt[3]{3}}{\sqrt[3]{6}}\cdot \cfrac{\sqrt[3]{6^2}}{\sqrt[3]{6^2}}\implies \cfrac{(2+\sqrt[3]{3})(\sqrt[3]{6^2})}{(\sqrt[3]{6})(\sqrt[3]{6^2})}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3}\cdot \sqrt[3]{36}}{\sqrt[3]{6^3}} \\\\\\ \cfrac{2\sqrt[3]{36}+\sqrt[3]{3\cdot 36}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{108}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3^3\cdot 4}}{6} \\\\\\ \cfrac{2\sqrt[3]{36}+3\sqrt[3]{ 4}}{6}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B2%2B%5Csqrt%5B3%5D%7B3%7D%7D%7B%5Csqrt%5B3%5D%7B6%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%5B3%5D%7B6%5E2%7D%7D%7B%5Csqrt%5B3%5D%7B6%5E2%7D%7D%5Cimplies%20%5Ccfrac%7B%282%2B%5Csqrt%5B3%5D%7B3%7D%29%28%5Csqrt%5B3%5D%7B6%5E2%7D%29%7D%7B%28%5Csqrt%5B3%5D%7B6%7D%29%28%5Csqrt%5B3%5D%7B6%5E2%7D%29%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B%5Csqrt%5B3%5D%7B3%7D%5Ccdot%20%5Csqrt%5B3%5D%7B36%7D%7D%7B%5Csqrt%5B3%5D%7B6%5E3%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B%5Csqrt%5B3%5D%7B3%5Ccdot%2036%7D%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B%5Csqrt%5B3%5D%7B108%7D%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B%5Csqrt%5B3%5D%7B3%5E3%5Ccdot%204%7D%7D%7B6%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B3%5Csqrt%5B3%5D%7B%204%7D%7D%7B6%7D)
so
