From Förberedande kurs i matematik 1

If we divide up \displaystyle \text{153 } and \displaystyle \text{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} & \text{153 }=3\centerdot 51=3\centerdot 3\centerdot 17=3^{2}\centerdot 17, \\ & 68=2\centerdot 37=2\centerdot 2\centerdot 17=2^{2}\centerdot 17 \\ \end{align}

We get

\displaystyle \begin{align} & \sqrt{\text{153}}\text{-}\sqrt{68}\text{ }=\sqrt{3^{2}\centerdot 17}-\sqrt{2^{2}\centerdot 17} \\ & =3\sqrt{17}-2\sqrt{17}=\sqrt{17} \\ \end{align}

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