Question

hey ( : can you help me simplify this radical expression? (in attached photo)

Answer 1

multiply top and bottom times \(\normalsize\color{black}{ \rm \sqrt[3]{x^2} }\) .

Answer 2

I don't think I understand how to do that.

Answer 3

\(\large \bf \cfrac{\sqrt[3]{24x^3}}{\sqrt[3]{x^3}}\qquad or \cfrac{\sqrt[3]{24x}}{\sqrt[3]{-x}} \quad ?\)

Answer 4

your two postings seem to differ some

Answer 5

actually top and bottom times, \(\normalsize\color{black}{ \rm \sqrt[3]{-x^2} }\)

Answer 6

the negative x one , sorry

Answer 7

\(\large {
\bf \cfrac{\sqrt[3]{24x}}{\sqrt[3]{-x}}\implies \cfrac{\sqrt[3]{24x}}{\sqrt[3]{-1\cdot x}}\implies \cfrac{\sqrt[3]{24x}}{\sqrt[3]{x}\cdot \sqrt[3]{-1}}
\\ \quad \\
{\color{blue}{ (-1)^3\to -1\cdot -1\cdot -1\to }}-1\qquad thus
\\ \quad \\
\cfrac{\sqrt[3]{24x}}{\sqrt[3]{x}\cdot \sqrt[3]{(-1)^3}}\implies \cfrac{\sqrt[3]{24x}}{-\sqrt[3]{x}}
\\ \quad \\
\sqrt[{\color{red} m}]{a^{\color{blue} n}}=a^{\frac{{\color{blue} n}}{{\color{red} m}}}\qquad thus
\\\quad \\
\cfrac{\sqrt[3]{24x}}{-\sqrt[3]{x}}\implies \cfrac{(24x)^{\frac{1}{3}}}{-(x)^{\frac{1}{3}}}
\\ \quad \\
\cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}= \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}}\implies a^{-\frac{{\color{blue} n}}{{\color{red} m}}} thus
\\\quad \\
\cfrac{(24x)^{\frac{1}{3}}}{-(x)^{\frac{1}{3}}}\implies \cfrac{(24x)^{\frac{1}{3}}}{-1}\cdot (x)^{-\frac{1}{3}}\implies
-(24x)^{\frac{1}{3}}\cdot (x)^{-\frac{1}{3}}
\\ \quad \\
-(24)^{\frac{1}{3}}\cdot x^{\frac{1}{3}}\cdot x^{-\frac{1}{3}}\implies -(24)^{\frac{1}{3}}\cdot x^{\frac{1}{3}-\frac{1}{3}}
} \)

Answer 8

keep in mind that \(\large \bf { 24\to 2\cdot 2\cdot 2\cdot 3\to 2^3\cdot 3\qquad thus
\\ \quad \\
-(24)^{\frac{1}{3}}\implies -\sqrt[3]{24}\implies -\sqrt[3]{2^3\cdot 3}\implies -\sqrt[3]{2^3}\cdot \sqrt[3]{3} }\)