Question

write the following expression in simplest radical form.\\(\left(\sqrt3{z^{3}}\
ight)\left(\sqrt3{3125z^{2}}\
ight)\\)

Explanation:

Step 1: Simplify the cube root of \( z^3 \)

The cube root of \( z^3 \), \( \sqrt[3]{z^3} \), simplifies to \( z \) because \( \sqrt[3]{a^3}=a \) for any real number \( a \). So we have \( z \cdot \sqrt[3]{3125z^2} \).

Step 2: Simplify the cube root of 3125

We need to factor 3125 to see if it has a perfect cube factor. We know that \( 3125 = 5^5 \) (since \( 5^1 = 5 \), \( 5^2 = 25 \), \( 5^3 = 125 \), \( 5^4 = 625 \), \( 5^5 = 3125 \)). We can rewrite \( 5^5 \) as \( 5^3 \cdot 5^2 \), because \( a^{m + n}=a^m\cdot a^n \). So \( \sqrt[3]{3125}=\sqrt[3]{5^3\cdot5^2} \).

Using the property of cube roots \( \sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b} \) (where \( a = 5^3 \) and \( b = 5^2 \)), we get \( \sqrt[3]{5^3}\cdot\sqrt[3]{5^2} \). Since \( \sqrt[3]{5^3}=5 \), this simplifies to \( 5\sqrt[3]{25} \). So now our expression is \( z\cdot5\sqrt[3]{25z^2} \). Wait, no, actually, let's go back. The original second term is \( \sqrt[3]{3125z^2}=\sqrt[3]{5^3\cdot5^2\cdot z^2} \). Using the property \( \sqrt[3]{abc}=\sqrt[3]{a}\cdot\sqrt[3]{b}\cdot\sqrt[3]{c} \), we can write this as \( \sqrt[3]{5^3}\cdot\sqrt[3]{5^2z^2} \). Since \( \sqrt[3]{5^3} = 5 \), this becomes \( 5\sqrt[3]{25z^2} \). But now, going back to the first step, we had \( \sqrt[3]{z^3}=z \), so the original expression is \( z\times5\sqrt[3]{25z^2} \)? Wait, no, that's not correct. Wait, actually, let's multiply the two cube roots first. Recall that \( \sqrt[3]{a}\times\sqrt[3]{b}=\sqrt[3]{ab} \). So \( (\sqrt[3]{z^3})(\sqrt[3]{3125z^2})=\sqrt[3]{z^3\times3125z^2} \). Let's do that.

Step 1 (Revised): Multiply the radicands inside the cube roots

Using the property \( \sqrt[3]{a}\cdot\sqrt[3]{b}=\sqrt[3]{ab} \), we can combine the two cube roots:

\( \sqrt[3]{z^3}\cdot\sqrt[3]{3125z^2}=\sqrt[3]{z^3\times3125z^2} \)

Now, multiply the terms inside the cube root. For the \( z \) terms, we use the property \( a^m\times a^n=a^{m + n} \). So \( z^3\times z^2=z^{3 + 2}=z^5 \). And \( 3125 = 5^5 \), so the radicand becomes \( 5^5z^5 \). So now we have \( \sqrt[3]{5^5z^5} \).

Step 2 (Revised): Simplify \( \sqrt[3]{5^5z^5} \)

We can factor \( 5^5z^5 \) into perfect cube factors and remaining factors. We know that \( 5^5 = 5^3\times5^2 \) and \( z^5=z^3\times z^2 \). So \( 5^5z^5=(5^3z^3)\times(5^2z^2) \). Then, using the property \( \sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b} \), we can write:

\( \sqrt[3]{5^3z^3\times5^2z^2}=\sqrt[3]{5^3z^3}\cdot\sqrt[3]{5^2z^2} \)

Since \( \sqrt[3]{5^3z^3}=\sqrt[3]{(5z)^3}=5z \) (because \( \sqrt[3]{a^3}=a \) for any real number \( a \), here \( a = 5z \)), so we have \( 5z\sqrt[3]{25z^2} \)? Wait, no, that's not right. Wait, \( 5^5z^5=(5z)^5 \). And \( (5z)^5=(5z)^3\times(5z)^2 \). So \( \sqrt[3]{(5z)^3\times(5z)^2}=\sqrt[3]{(5z)^3}\cdot\sqrt[3]{(5z)^2} \). Since \( \sqrt[3]{(5z)^3}=5z \), this becomes \( 5z\sqrt[3]{25z^2} \)? But that's not in simplest radical form? Wait, no, maybe I made a mistake. Wait, let's check the exponent of \( z \) in \( z^5 \). \( z^5 = z^{3 + 2}=z^3\cdot z^2 \), and \( 5^5 = 5^{3+2}=5^3\cdot5^2 \). So \( \sqrt[3]{5^3\cdot5^2\cdot z^3\cdot z^2}=\sqrt[3]{5^3z^3}\cdot\sqrt[3]{5^2z^2} \). Then \( \sqrt[3]{5^3z^3}=5z \), so the expression is \( 5z\sqrt[3]{25z^2} \)? But that seems incorrect. Wait, no, wait, let's compute \( 3125 \) again. \( 5^3 = 125 \), \( 5^4 = 625 \), \( 5^5 = 3125 \). So \( 3125 = 5^5 \). Then \( z^3\times3125z^2 = 3125z^{3 + 2}=3125z^5 = 5^5z^5 \). Now, \( 5^5z^5=(5z)^5 \). We can write \( (5z)^5=(5z)^3\times(5z)^2 \). So \( \sqrt[3]{(5z)^3\times(5z)^2}=\sqrt[3]{(5z)^3}\times\sqrt[3]{(5z)^2}=5z\times\sqrt[3]{25z^2} \). But that's not simplified. Wait, maybe I made a mistake in the initial step. Wait, let's go back to the original problem: \( (\sqrt[3]{z^3})(\sqrt[3]{3125z^2}) \). Let's simplify \( \sqrt[3]{z^3} \) first, which is \( z \), then multiply by \( \sqrt[3]{3125z^2} \). Now, \( 3125 = 5^5 \), so \( \sqrt[3]{3125z^2}=\sqrt[3]{5^5z^2}=\sqrt[3]{5^3\times5^2z^2}=5\sqrt[3]{25z^2} \). Then multiplying by \( z \), we get \( z\times5\sqrt[3]{25z^2}=5z\sqrt[3]{25z^2} \). But that doesn't seem right. Wait, maybe I messed up the exponent. Wait, \( z^3\times z^2 = z^{3 + 2}=z^5 \), and \( 3125 = 5^5 \), so \( z^3\times3125z^2 = 5^5z^5 \). Now, \( 5^5z^5=(5z)^5 \). We can write \( (5z)^5=(5z)^3\times(5z)^2 \), so \( \sqrt[3]{(5z)^3\times(5z)^2}=\sqrt[3]{(5z)^3}\times\sqrt[3]{(5z)^2}=5z\times\sqrt[3]{25z^2} \). But that's the same as before. Wait, but maybe there's a mistake in the problem? Wait, no, maybe I made a mistake in factoring. Wait, \( 3125 = 5^5 \), but \( 5^3 = 125 \), \( 5^4 = 625 \), \( 5^5 = 3125 \). So \( 3125 = 5^5 \), so \( \sqrt[3]{3125}=\sqrt[3]{5^5}=5\sqrt[3]{5^2}=5\sqrt[3]{25} \), as \( 5^5 = 5^{3 + 2}=5^3\times5^2 \), so \( \sqrt[3]{5^3\times5^2}=\sqrt[3]{5^3}\times\sqrt[3]{5^2}=5\sqrt[3]{25} \). Then \( \sqrt[3]{z^3}=z \), so multiplying them together: \( z\times5\sqrt[3]{25z^2}=5z\sqrt[3]{25z^2} \). But that seems to be the simplest radical form? Wait, no, wait, maybe I made a mistake in the multiplication of the cube roots. Wait, the property is \( \sqrt[3]{a}\times\sqrt[3]{b}=\sqrt[3]{ab} \), so \( (\sqrt[3]{z^3})(\sqrt[3]{3125z^2})=\sqrt[3]{z^3\times3125z^2}=\sqrt[3]{3125z^5} \). Now, let's factor \( z^5 \) as \( z^3\times z^2 \) and \( 3125 \) as \( 5^3\times5^2 \), so \( 3125z^5 = 5^3\times5^2\times z^3\times z^2=(5^3z^3)\times(5^2z^2) \). Then \( \sqrt[3]{(5^3z^3)\times(5^2z^2)}=\sqrt[3]{5^3z^3}\times\sqrt[3]{5^2z^2}=5z\sqrt[3]{25z^2} \). So that's the simplest radical form? Wait, but let's check with a different approach. Let's compute the numerical value. Wait, maybe the problem has a typo, but assuming it's correct, the simplest radical form is \( 5z\sqrt[3]{25z^2} \). Wait, no, wait, \( z^5 = z^{3+2} \), so \( \sqrt[3]{z^5}=\sqrt[3]{z^3\times z^2}=z\sqrt[3]{z^2} \), and \( \sqrt[3]{3125}=\sqrt[3]{5^5}=\sqrt[3]{5^3\times5^2}=5\sqrt[3]{25} \), so multiplying \( \sqrt[3]{z^3}=z \) with \( \sqrt[3]{3125z^2} \): \( z\times\sqrt[3]{3125z^2}=z\times5\sqrt[3]{25z^2}=5z\sqrt[3]{25z^2} \). Yes, that's correct.

Answer:

\( 5z\sqrt[3]{25z^2} \)