Question

Solve the following radical expression: ∛3/∛16
Please show complete solution and answer

Answer 1

@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?

Answer 2

@OffnenStudieren the correct answer is (3x∛4xy)/2y but idk how to get it

Answer 3

Is this doubt cleared?

Answer 4

@AkashdeepDeb not yet, i still don't know how that became the answer

Answer 5

You're referring to the wrong answering key or something then.
There IS NO y or x !! :P

Answer 6

@AkashdeepDeb oh sorry! the right answer is ∛12/4

Answer 7

The actual answer should be
∛3/16 :)
Check the question once more.

Answer 8

@AkashdeepDeb how'd you get that?

Answer 9

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
:)

Answer 10

@AkashdeepDeb but don't you have to remove the radical in the denominator?

Answer 11

\[\frac{ \sqrt[3]{3} }{ \sqrt[3]{16} }= \frac{ \sqrt[3]{3} }{ \sqrt[3]{2 \cdot 2 \cdot 2 \cdot 2}} =\frac{ \sqrt[3]{3} }{ 2\sqrt[3]{2 } }\]
if we "rationalize" by multiplying top and bottom by the cube root of 2 squared
we get
\[ \frac{ \sqrt[3]{3} }{ 2 \sqrt[3]{2}} \cdot \frac{ \sqrt[3]{2}\sqrt[3]{2}}{ \sqrt[3]{2}\sqrt[3]{2}}= \frac{ \sqrt[3]{3}\cdot \sqrt[3]{4} }{ 2\cdot 2 } \\
= \frac{ \sqrt[3]{12} }{ 4}
\]

Answer 12

@phi wait, how did you 2∛2 become ∛2∛2 when you rationalized it?

Answer 13

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

Answer 14

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 \( \sqrt[3]{2} \)

Answer 15

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
\[ 2^{\frac{1}{3}}2^{\frac{1}{3}}2^{\frac{1}{3} } = 2^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=2^1 = 1\]

Answer 16

@phi Do you know a quick proof that $$c=\frac{c}{1}$$

Answer 17

@skullpatrol you would have to go back to the axioms and definitions of your group and its defined operations.

Answer 18

***fix typo at the very end
\[ 2^{\frac{1}{3}}2^{\frac{1}{3}}2^{\frac{1}{3} } = 2^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=2^1 = 2 \]