+1678 Rationalize the denominator of $\displaystyle \frac{1}{2 - \sqrt[3]{2}}$. 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 $A + B + C + D$?
+177 To rationalize an expression such as \(\frac{1}{2 - \sqrt[3]{2}}\), you first recognize that there is a cube root in the denominator. This immediately provides insight as to how to approach this particular problem. Personally, I find working with fractional exponents is far easier than radicals, so I will almost always convert to this.
Since there is a cube root within the denominator, we will have to make the denominator a perfect cube in order to ensure that the denominator has no radicals. We can do this by making a key observation: \(A^3 - B^3 = (A - B)(A^2 + AB + B^2)\). This is the formula for the difference of a perfect cube. This information should be enough to rationalize this problem and ones like it.
If we look closely at the denominator, we can match up the formula for a perfect cube to the denominator of this particular problem.
If we let \(A = 2\) and \(B = \sqrt[3]{2} = 2^\frac{1}{3}\), then that would make the following identity.
\(A = 2; B = 2^\frac{1}{3} \\ 2^3 - \left(2^\frac{1}{3}\right)^3 = \left(2 - 2^\frac{1}{3}\right)\left(2^2 + 2*2^\frac{1}{3} + \left(2^\frac{1}{3}\right)^2\right) \\ 6 = \left(2 - 2^\frac{1}{3} \right) \left(4 + 2^\frac{4}{3} + 2^\frac{2}{3}\right) \\\)
Now, we can rationalize because we have the term that will cancel out all radicals within the denominator.
\(\frac{1}{2 - 2^\frac{1}{3}} * \frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{4 + 2^\frac{4}{3} + 2 ^\frac{2}{3}} = \frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{6}\)
Now, we have to convert to the radical notation and figure out how it matches the desired format of \(\displaystyle \frac{\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C}}{D}\).
\(\frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{6} = \frac{4 + \sqrt[3]{2^4} + \sqrt[3]{2^2}}{6} = \frac{\sqrt[3]{64} + \sqrt[3]{16} + \sqrt[3]{4}}{6}\)
Now, we identify the desired variables and find their sum.
\(A = 64, B=16, C=4, D=6 \\ A + B + C + D = 64 + 16 + 4 + 6 = 90\)
+177 To rationalize an expression such as \(\frac{1}{2 - \sqrt[3]{2}}\), you first recognize that there is a cube root in the denominator. This immediately provides insight as to how to approach this particular problem. Personally, I find working with fractional exponents is far easier than radicals, so I will almost always convert to this.
Since there is a cube root within the denominator, we will have to make the denominator a perfect cube in order to ensure that the denominator has no radicals. We can do this by making a key observation: \(A^3 - B^3 = (A - B)(A^2 + AB + B^2)\). This is the formula for the difference of a perfect cube. This information should be enough to rationalize this problem and ones like it.
If we look closely at the denominator, we can match up the formula for a perfect cube to the denominator of this particular problem.
If we let \(A = 2\) and \(B = \sqrt[3]{2} = 2^\frac{1}{3}\), then that would make the following identity.
\(A = 2; B = 2^\frac{1}{3} \\ 2^3 - \left(2^\frac{1}{3}\right)^3 = \left(2 - 2^\frac{1}{3}\right)\left(2^2 + 2*2^\frac{1}{3} + \left(2^\frac{1}{3}\right)^2\right) \\ 6 = \left(2 - 2^\frac{1}{3} \right) \left(4 + 2^\frac{4}{3} + 2^\frac{2}{3}\right) \\\)
Now, we can rationalize because we have the term that will cancel out all radicals within the denominator.
\(\frac{1}{2 - 2^\frac{1}{3}} * \frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{4 + 2^\frac{4}{3} + 2 ^\frac{2}{3}} = \frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{6}\)
Now, we have to convert to the radical notation and figure out how it matches the desired format of \(\displaystyle \frac{\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C}}{D}\).
\(\frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{6} = \frac{4 + \sqrt[3]{2^4} + \sqrt[3]{2^2}}{6} = \frac{\sqrt[3]{64} + \sqrt[3]{16} + \sqrt[3]{4}}{6}\)
Now, we identify the desired variables and find their sum.
\(A = 64, B=16, C=4, D=6 \\ A + B + C + D = 64 + 16 + 4 + 6 = 90\)
