Question

(6-3(cube root of 6)/(cube root of 9)

Answer 1

For this case we must simplify the following expression:

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

Multiplying the numerator and denominator by$(\sqrt [3] {9}) ^ 2$

$\frac {6-3 \sqrt [3] {6}} {\sqrt [3] {9}} * \frac {(\sqrt [3] {9}) ^ 2} {(\sqrt [3] { 9}) ^ 2} =$

We rewrite:

$\frac {\frac {6-3 \sqrt [3] {6}} * (\sqrt [3] {9}) ^ 2} {\sqrt [3] {9} * (\sqrt [3] {9 }) ^ 2} =$

By properties of powers we have that:

$a ^ m * a ^ n = a ^ {m + n}\\\frac {(6-3 \sqrt [3] {6}) * (\sqrt [3] {9}) ^ 2} {(\sqrt [3] {9}) ^ 3} =\\\frac {(6-3 \sqrt [3] {6}) * (\sqrt [3] {9}) ^ 2} {9} =$

We rewrite, moving the exponent within the radical:

$\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {9 ^ 2}} {9} =\\\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {81}} {9} =$

We can rewrite$3 * 3 ^ 3 = 81$

$\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {3 * 3 ^ 3}} {9} =$

We simplify:

$\frac {(6-3 \sqrt [3] {6}) * 3 \sqrt [3] {3}} {9} =$

We apply distributive property:

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

Simplifying we finally have:

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

Answer:

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