From Förberedande kurs i matematik 1

If we divide up 153 and 68 into their smallest possible integer factors, we can see whether there are squared numbers which can be taken out from under the roots, and if the expression can be simplified further,

\displaystyle \begin{align} 153 &= 3\cdot 51 = 3\cdot 3\cdot 17 = 3^{2}\cdot 17,\\[5pt] 68 &= 2\cdot 37 = 2\cdot 2\cdot 17 = 2^{2}\cdot 17\,\textrm{.} \end{align}

We get

\displaystyle \begin{align} \sqrt{153}-\sqrt{68} &= \sqrt{3^{2}\cdot 17}-\sqrt{2^{2}\cdot 17}\\[5pt] &= 3\sqrt{17}-2\sqrt{17}\\[5pt] &= \sqrt{17}\,\textrm{.} \end{align}

Note: A good tip for when we investigate whether an integer is divisible by 3 is to look at the sum of the number's digits. A number is divisible by 3 if and only if the sum of its digits is divisible by 3. E.g. the number 97818 is divisible by 3 because the sum of its digits, \displaystyle 9+7+8+1+8=33, is divisible by 3 and, conversely, the number 11536 is not divisible by 3 since sum of its digits, \displaystyle 1+1+5+3+6=16, is not divisible by 3.