Rationalize the denominator of \(\displaystyle \frac{1}{\sqrt[3]{3} - \sqrt[3]{2}}\) . With your answer in the form \(\displaystyle \frac{\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C}}{D}\) , and the fraction in lowest terms, what is A+B+C+D?

tertre
May 29, 2017

#1**+2 **

We can write

1 / [ 3^(1/3) - 2 ^(1/3) ]

Multiply top/bottom by [ 3^(2/3) + (3*2)^(1/3) + 2^(2/3) ]

And we have

[ 3^(2/3) + (3*2)^(1/3) + 2^(2/3) ] / [ 3 + (18)^(1/3) + (12)^(1/3) - (18)^(1/3) - 12^(1/3) - 2 ]

[ (9)^(1/3) + (6)^(1/3) + (4)^(1/3) ] / 1

So A + B + C + D = 9 + 6 + 4 + 1 = 20

CPhill
May 29, 2017