Question 464643
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The other tutor's solution is wrong.

Prime factor them all:

21a²b&#8310; = 3·7·a·a·b·b·b·b·b·b

3a<sup><font size = 1>4</font></sup>b&#8312; = 3·a·a·a·a·b·b·b·b·b·b·b·b

14a&#8309;b&#8312; = 2·7·a·a·a·a·a·b·b·b·b·b·b·b·b

 Line those up like this so that only like factors
are lined up vertically. Then draw a line underneath:


  3·7·a·a      ·b·b·b·b·b·b
  3  ·a·a·a·a  ·b·b·b·b·b·b·b·b
2  ·7·a·a·a·a·a·b·b·b·b·b·b·b·b
-------------------------------

Bring every factor down on the bottom line

  3·7·a·a      ·b·b·b·b·b·b
  3  ·a·a·a·a  ·b·b·b·b·b·b·b·b
2  ·7·a·a·a·a·a·b·b·b·b·b·b·b·b
-------------------------------
2·3·7·a·a·a·a·a·b·b·b·b·b·b·b·b

Multiply all that together and you get:


    42a&#8309;b&#8312;

A shorter way is:

1.  Get the LCM of just the coefficients 21,3,and 14, which is 42
2.  use the largest power of each letter that occurs in any one 
    factor.  That's a&#8309; and b&#8312;, for they are the largest
    powers of a and b that appear in any of the 3 original expressions.

Edwin</pre>