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In $3-41$ , express each product in simplest form. Variables in the radicand with an even index are non-negative.$$\sqrt[3]{15 a^{2}} \cdot \sqrt[3]{9 a^{4}}$$

$3 a^{2} \sqrt[3]{5}$

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Chapter 3

REAL NUMBERS AND RADICALS

Section 5

Multiplying Radicals

Whole which of Numbers

Fractions and Mixed Numbers

Decimals

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probably or solving for the cubed root of 15 a squared times the cubed root of 98 to the fourth. To do so, we need to keep in mind that if we have thes same radicals So, for example, the cubed root and the radicals air being multiplied together, we can just multiply their insides in one overall radical sign. What I mean by that is we can just rewrite this as Thedc. You'd route of 15 a squared times nine a to the fourth. Continuing to simplify, I'm gonna multiply these whole numbers together the 15 and the nine to get 135. So I think you'd of 135 times a to the two plus four. When we multiply variables together, we add their exponents. This gives me the cubed root of 135 times a to the sixth. The next step takes a little bit thought and try on air to figure out what numbers could possibly multiply together to get 135. And we ideally want numbers where one of them is a perfect que because we're dealing with a cube root here So if I think about it, I can realize that 27 times five is 135 and 27 itself is a perfect cube. The cubed root of 27 is three. So rewriting this, I'm gonna have the cubed root off 27 times, five times a to the sixth. I can rewrite it because 27 times five is 135. So this is a perfectly valid way of writing it. And now I'm gonna break apart the individual pieces who now have the cubed root of 27 times the cubed root of five times three cubed root of a to the sixth. For finding our final answer. We're just gonna simplify each part. The cubed root of 27 is three. The cubed root of five is just cute of fives was B times the huge route of five. And then we need to find the cubed root of a to the sixth. The cube root of eight of the sixth is the same thing as a to the sixth to the 1/3 power. And when we write it like that, we could just multiply thes exponents together because we have an exponent race to another exponents, we multiply. So this would be equal to a 2 6/3 just just equal to a to the second power. So this term, right here the cubed root of aid to the sixth, we can just rewrite as a squared. So our final answer if we re arrange these in the correct order is three our whole number a squared the variable turns the cubed root of five three, a squared times a Queudrue defy It's our final answer.

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