Question 819768
Since 1/3 as an exponent means cube root and since radicals display better on algebra.com than fractional exponents, I am going to rewrite the expression with radicals. (Note: This is not required. All the steps below work out exactly the same with exponents of 1/3.)<br>
{{{14/((5^(1/3))-(2^(1/3)))}}}
{{{14/(root(3, 5)-root(3, 2))))<br>
With the cube roots in the denominator we want to find a way to change the denominator so that there are only perfect cube terms. A factoring pattern is useful here:
{{{(a-b)(a^2+ab+b^2) = a^3-b^3}}}
One way to look at this is that it shows use how to take two terms with a "-" between them, multiply it by something, and end up with an expression of nothing but perfect cubes.<br>
Our denominator is two terms with a "-" between them. So if we treat the {{{root(3, 5)}}} as the "a" and the {{{root(3, 2)}}} as the "b" in the pattern, the pattern tells what what to multiply it by to get perfect cubes: {{{(root(3, 5))^2+(root(3, 5))(root(3, 2))+(root(3, 2))^2}}}. But we can't just multiply the denominator by something. We must multiply the numerator by the same thing, too:
{{{(14/(root(3, 5)-root(3, 2)))(((root(3, 5))^2+(root(3, 5))(root(3, 2))+(root(3, 2))^2)/((root(3, 5))^2+(root(3, 5))(root(3, 2))+(root(3, 2))^2)))))}}}
Before we actually multiply, let's simplify the second fraction:
{{{(14/(root(3, 5)-root(3, 2)))((root(3, 25)+root(3, 10)+root(3, 4))/(root(3, 25)+root(3, 10)+root(3, 4)))}}}
Now we multiply. In the numerator we use the Distributive Property. In the denominator the pattern tells us how it works out:
{{{(14root(3, 25)+14root(3, 10)+14root(3, 4))/((root(3, 5))^3-(root(3, 2))^3)}}}
which simplifies as follows:
{{{(14root(3, 25)+14root(3, 10)+14root(3, 4))/(5-2)}}}
{{{(14root(3, 25)+14root(3, 10)+14root(3, 4))/3}}}