Given Expression: We are given the expression: root(3)(v) * sqrt(v) First, we need to express both roots in terms of exponents. The cube root of v can be written as v^(1/3) and the square root of v can be written as v^(1/2).Express Roots as Exponents: Now we multiply the two expressions using the property of exponents that states when we multiply like bases, we add the exponents. v^(1/3) * v^(1/2) = v^(1/3 + 1/2)Multiply Expressions with Exponents: To add the exponents, we need a common denominator. The common denominator of 3 and 2 is 6. So we convert 1/3 and 1/2 to have the denominator of 6. 1/3 = 2/6 and 1/2 = 3/6Add Exponents with Common Denominator: Now we add the exponents with the common denominator. v^(2/6 + 3/6) = v^(5/6)Simplest Form of Expression: The expression v^(5/6) is the simplest form of the product of the cube root of v and the square root of v.

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$\sqrt[3]{v} \cdot \sqrt{v}$ =

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Algebra 1

Multiply radical expressions

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\sqrt[3]{v} \cdot \sqrt{v}

Given Expression: We are given the expression: $\sqrt[3]{v} \times \sqrt{v}$ $\newline$ First, we need to express both roots in terms of exponents. $\newline$ The cube root of $v$ can be written as $v^{1/3}$ and the square root of $v$ can be written as $v^{1/2}$ .
Express Roots as Exponents: Now we multiply the two expressions using the property of exponents that states when we multiply like bases, we add the exponents. $\newline$ $v^{\frac{1}{3}} \times v^{\frac{1}{2}} = v^{\frac{1}{3} + \frac{1}{2}}$
Multiply Expressions with Exponents: To add the exponents, we need a common denominator. The common denominator of $3$ and $2$ is $6$ . So we convert $\frac{1}{3}$ and $\frac{1}{2}$ to have the denominator of $6$ . $\frac{1}{3} = \frac{2}{6}$ and $\frac{1}{2} = \frac{3}{6}$
Add Exponents with Common Denominator: Now we add the exponents with the common denominator. $\newline$ $v^{\frac{2}{6} + \frac{3}{6}} = v^{\frac{5}{6}}$
Simplest Form of Expression: The expression $v^{5/6}$ is the simplest form of the product of the cube root of $v$ and the square root of $v$ .