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# Algebra

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Rationalize the denominator of $\frac{1}{\sqrt[3]{2} + \sqrt{16}}$

Nov 29, 2021

#1
+155
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$${ \ {1} \over 3 root 2 +4}\\$$

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Nov 29, 2021
#2
+69
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Actualy rationalizing the denominator means getting rid of the roots, you did not do that.

Nov 29, 2021
#3
+115901
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The sum of 2 cubics is  ..  (something you need to memorize)

$$a^3+b^3=(a+b)(a^2-ab+b^2)$$

So if you have a denominator of  $$(a+b)$$ where a or b have cube roots, and you multiply it by  $$(a^2-ab+b^2)$$ then you will end up with a rational denominator.

$$\frac{1}{\sqrt[3]{2}+\sqrt{16}}=\frac{1}{\sqrt[3]{2}+4}$$

let

$$a= 2^{1/3}\qquad and\qquad b=4\\so\\ a^2=2^{2/3}\qquad and\qquad b^2=16\\ ab=4*2^{1/3}$$

$$a^3+b^3=(a+b)(a^2-ab+b^2)$$

$$\frac{1}{\sqrt[3]{2}+4}\times \frac{2^{2/3}-4*2^{(1/3)}+4^2}{2^{2/3}-4*2^{(1/3)}+4^2}\\~\\ =\frac{2^{2/3}-4*2^{(1/3)}+4^2}{2+64}\\~\\ =\frac{\sqrt[3]{4}-4\sqrt[3]{2}+16}{66}\\~\\$$