Question

What is the following quotient 6-3(^3√6)/^3√9

Answer 1

Answer:

$2\sqrt[3]{3}-\sqrt[3]{18}$

Step-by-step explanation:

We first rationalize the denominator.  This means we multiply the numerator and denominator by a factor that will make the denominator a whole number.

Our denominator is $\sqrt[3]{9}$; this can also be written as $9^{\frac{1}{3}}$.

For 9 to be a whole number, its exponent must be a whole number.  We want to add 2/3 to the exponent of 9 on the bottom; this means we multiply by $9^{\frac{2}{3}}$:

On the numerator, this gives us:

6(9^(2/3))-3(6^(1/3))(9^(2/3))

= $6\sqrt[3]{9(9)} -3\sqrt[3]{2(3)}(\sqrt[3]{9(9)}) \\\\=6\sqrt[3]{3(3)(3)(3)}-3\sqrt[3]{2(3)(3)(3)(3)(3)}\\\\=6(3)\sqrt[3]{3}-3(3)\sqrt[3]{2(3)(3)}\\\\=18\sqrt[3]{3}-9\sqrt[3]{18}$

On the denominator, we will have:

(9^(1/3))(9^(2/3)) = 9^(3/3) = 9^1 = 9

This gives us:

$\frac{18\sqrt[3]{3}-9\sqrt[3]{18}}{9}\\\\=2\sqrt[3]{3}-\sqrt[3]{18}$

Answer 2

Quotient given:

$\frac{6-3 \sqrt[3]{6} }{ \sqrt[3]{9} }$

Answer:

$2 \sqrt[3]{3} - \sqrt[3]{18}$

Explanation:

1) Rationalize by multiplying numerator and denominator by        $\sqrt[3]{9}$

2) 
$\frac{6-3 \sqrt[3]{6} }{ \sqrt[3]{9} } \frac{\sqrt[3]{9}}{\sqrt[3]{9}}$

3) Due the operations

$\frac{6 \sqrt[3]{3}-3 \sqrt[3]{6.3}}{ \sqrt[3]{27} }$

$\frac{6 \sqrt[3]{3}-3 \sqrt[3]{6.3}}{ 3 }=2 \sqrt[3]{3} - \sqrt[3]{18}$