Question 36507
I am going to assume you mean (2a+2b+2c)/(2(pi)a+2(pi)b+2(pi)c), or in other words, {{{(2a+2b+2c)/(2*pi*a+2*pi*b+2*pi*c)}}}
If that is what you meant, first factor 2 from the three terms of the numerator and {{{2*pi}}} from the three terms of the denominator: {{{2*(a+b+c)/(2*pi(a+b+c))}}}.  Now cancel 2 and (a+b+c) to get the result: {{{1/pi}}}