Question 220109
{{{(6x^(-1/3)+2x^(5/3))/(2x^(-4/3))}}}
First let's reduce the fraction. Reduced fractions are usually easier to work with. Since 2 is a common factor we can reduce the fraction be a factor of 2:
{{{(6x^(-1/3)+2x^(5/3))/(2x^(-4/3)) = (2(3x^(-1/3)+x^(5/3)))/(2x^(-4/3)) = (3x^(-1/3)+x^(5/3))/(x^(-4/3))}}}<br>
From this point there are, as usual, a number of ways we can go. The most direct way to the answer is to multiply the numerator and denominator by {{{x^(4/3)}}}. (I hope you can see that this will turn the denominator into a 1 which, in turn, means that we will no longer have a fraction. And not having a fraction anymore is, I hope you agree, a good thing.):
{{{(3x^(-1/3)+x^(5/3))/(x^(-4/3)) = ((3x^(-1/3)+x^(5/3))/(x^(-4/3)))((x^(4/3))/(x^(4/3))) =
((3x^(-1/3)+x^(5/3))(x^(4/3)))/(x^(-4/3)*x^(4/3))}}}
In the denominator, since<ul><li>the rule for exponents when you multiply is to add the exponents, and</li><li>-4/3 + 4/3 = 0, and</li><li>{{{x^0 = 1}}}</li></ul>the denominator becomes a 1!
{{{((3x^(-1/3)+x^(5/3))(x^(4/3)))/(x^(-4/3)*x^(4/3)) =                                 ((3x^(-1/3)+x^(5/3))(x^(4/3)))/1 = (3x^(-1/3)+x^(5/3))(x^(4/3))}}}
Now let's simplify using the Distributive Property:
{{{(3x^(-1/3)+x^(5/3))(x^(4/3)) = 3x^(-1/3)*x^(4/3)+x^(5/3)x^(4/3)}}}
Again we'll use the "add the exponents when multiplying" rule:
{{{3x^(-1/3)*x^(4/3)+x^(5/3)x^(4/3) = 3x^(3/3)+x^(9/3) = 3x^1 + x^3 = 3x + x^3}}}
And we're finished!