Radicals

**mr_quick** · May 13th 2009, 12:15 PM

3√[24x^5 y^3]
3 is the index (the small number in top of the radical sign)
Please help me I do Appreciate it

**Plato** · May 13th 2009, 12:21 PM

$\sqrt[3]{{24x^5 y^3 }} = 2xy\sqrt[3]{{3x^2 }}$

**mr_quick** · May 13th 2009, 12:24 PM

Originally Posted by Plato

$\sqrt[3]{{24x^5 y^3 }} = 2xy\sqrt[3]{{3x^2 }}$

Thanks for the help and if you can could you show me the solution

**masters** · May 13th 2009, 12:24 PM

Originally Posted by mr_quick

3√[24x^5 y^3]
3 is the index (the small number in top of the radical sign)
Please help me I do Appreciate it

Hi mr_quick,

$\sqrt[3]{24x^5y^3}=\sqrt[3]{2^3\cdot 3 \cdot x^3 \cdot x^2 \cdot y^3}=2xy \sqrt[3]{3x^2}$

**artvandalay11** · May 13th 2009, 12:27 PM

$\sqrt[3]{24x^5y^3}=(24x^5y^3)^\frac{1}{3}=24^\frac{1}{3}(x ^5)^\frac{1}{3}(y^3)^\frac{1}{3}$ now there's an exponent rule that says to multiply powers together when you have powers raised to powers so that becomes $2\sqrt[3]{3}x^\frac{5}{3}y$ or $2y(3x^5)^\frac{1}{3}$

**mr_quick** · May 13th 2009, 12:28 PM

Originally Posted by artvandalay11

$\sqrt[3]{24x^5y^3}=(24x^5y^3)^\frac{1}{3}=24^\frac{1}{3}(x ^5)^\frac{1}{3}(y^3)^\frac{1}{3}$ now there's an exponent rule that says to multiply powers together when you have powers raised to powers so that becomes $2\sqrt[3]{3}x^\frac{5}{3}y$ or $2y(3x^5)^\frac{1}{3}$

Thank you for the help I need it

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