elleblythe
  • elleblythe
Solve the following radical expression: ∛3/∛16 Please show complete solution and answer
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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elleblythe
  • elleblythe
@OffnenStudieren yeah so it becomes 3^(1/3) right? so what's next after that? i then multiply the expression with the conjugate of the denominator?
elleblythe
  • elleblythe
@OffnenStudieren the correct answer is (3x∛4xy)/2y but idk how to get it
AkashdeepDeb
  • AkashdeepDeb
Is this doubt cleared?

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elleblythe
  • elleblythe
@AkashdeepDeb not yet, i still don't know how that became the answer
AkashdeepDeb
  • AkashdeepDeb
You're referring to the wrong answering key or something then. There IS NO y or x !! :P
elleblythe
  • elleblythe
@AkashdeepDeb oh sorry! the right answer is ∛12/4
AkashdeepDeb
  • AkashdeepDeb
The actual answer should be ∛3/16 :) Check the question once more.
elleblythe
  • elleblythe
@AkashdeepDeb how'd you get that?
AkashdeepDeb
  • AkashdeepDeb
Because the cube root is takne common for both terms! :) It is just like this.. a^2.b^2 = (ab)^2 So ∛3/∛16 = ∛3/16 :)
elleblythe
  • elleblythe
@AkashdeepDeb but don't you have to remove the radical in the denominator?
phi
  • phi
\[\frac{ \sqrt[3]{3} }{ \sqrt[3]{16} }= \frac{ \sqrt[3]{3} }{ \sqrt[3]{2 \cdot 2 \cdot 2 \cdot 2}} =\frac{ \sqrt[3]{3} }{ 2\sqrt[3]{2 } }\] if we "rationalize" by multiplying top and bottom by the cube root of 2 squared we get \[ \frac{ \sqrt[3]{3} }{ 2 \sqrt[3]{2}} \cdot \frac{ \sqrt[3]{2}\sqrt[3]{2}}{ \sqrt[3]{2}\sqrt[3]{2}}= \frac{ \sqrt[3]{3}\cdot \sqrt[3]{4} }{ 2\cdot 2 } \\ = \frac{ \sqrt[3]{12} }{ 4} \]
elleblythe
  • elleblythe
@phi wait, how did you 2∛2 become ∛2∛2 when you rationalized it?
phi
  • phi
It didn't. we multiply top and bottom by \[ \sqrt[3]{2}\sqrt[3]{2} \] the bottom becomes \[ 2 \sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}\] the cube root of 2 times itself 3 times gives us 2, and the bottom becomes 2*2 or 4
phi
  • phi
The idea is we multiply the fraction \[ \frac{ \sqrt[3]{3} }{ 2\sqrt[3]{2 } } \] by 1. But we pick a special form of 1: \[ \frac{ \sqrt[3]{2} \sqrt[3]{2} }{ \sqrt[3]{2} \sqrt[3]{2 } } \] because we want three cube roots multiplied together in the bottom to get rid of the \( \sqrt[3]{2} \)
phi
  • phi
If you are familiar with exponents, \[ \sqrt[3]{2} = 2^{\frac{1}{3} }\] and you can see \[ \sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}= 2^{\frac{1}{3}}2^{\frac{1}{3}}2^{\frac{1}{3} }\] when you multiply terms with the same base, you add exponents. you get \[ 2^{\frac{1}{3}}2^{\frac{1}{3}}2^{\frac{1}{3} } = 2^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=2^1 = 1\]
skullpatrol
  • skullpatrol
@phi Do you know a quick proof that $$c=\frac{c}{1}$$
phi
  • phi
@skullpatrol you would have to go back to the axioms and definitions of your group and its defined operations.
phi
  • phi
***fix typo at the very end \[ 2^{\frac{1}{3}}2^{\frac{1}{3}}2^{\frac{1}{3} } = 2^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=2^1 = 2 \]

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