Question 152399
Use the following properties of exponents:
(1)
{{{a^m*a^n=a^(m+n)}}}
(2)
{{{1/a^(-m)=a^m}}}
So
{{{(2n^(1/3)(3n^(1/3)-4n^(2/3)))/(2n^(-1/3))}}}
={{{(n^(1/3)(3n^(1/3)-4n^(2/3)))/n^(-1/3)}}}
={{{(n^(1/3)n^(1/3)(3n^(1/3)-4n^(2/3)))}}}
={{{(n^(1/3+1/3)(3n^(1/3)-4n^(2/3)))}}}
={{{(n^(2/3)(3n^(1/3)-4n^(2/3)))}}}
={{{n^(2/3)*3*n^(1/3)-n^(2/3)*4*n^(2/3)}}}
={{{3n^(2/3)n^(1/3)-4n^(2/3)n^(2/3)}}}
={{{3n^(2/3+1/3)-4n^(2/3+2/3)}}}
={{{3n^1-4n^(4/3)}}}
={{{3n^1-4n^(1+1/3)}}}
={{{3n-4n^1n^(1/3)}}}
={{{3n-4n*n^(1/3)}}}