Question

what is the following product? assume y ≥ 0. \\(\sqrt{y^3} \cdot \sqrt{y^3}\\) options: \\(2y^3\\), \\(y^6\\), \\(2y^6\\), \\(y^3\\)

Explanation:

Step1: Recall radical multiplication rule

For non - negative real numbers \(a\) and \(b\), \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\). So, \(\sqrt{y^{3}}\cdot\sqrt{y^{3}}=\sqrt{y^{3}\cdot y^{3}}\)

Step2: Simplify the radicand

Using the exponent rule \(a^{m}\cdot a^{n}=a^{m + n}\), for \(a = y\), \(m=3\) and \(n = 3\), we have \(y^{3}\cdot y^{3}=y^{3 + 3}=y^{6}\)

Step3: Simplify the square root

Since \(y\geq0\), \(\sqrt{y^{6}}=y^{3}\)? Wait, no, wait. Wait, the original problem is \(\sqrt{y^{3}}\cdot\sqrt{y^{3}}\), wait, maybe I misread. Wait, no, the problem is \(\sqrt{y^{3}}\cdot\sqrt{y^{3}}\)? Wait, no, looking at the image, the problem is \(\sqrt{y^{3}}\cdot\sqrt{y^{3}}\)? Wait, no, maybe it's a typo? Wait, no, let's re - examine. Wait, the problem is \(\sqrt{y^{3}}\cdot\sqrt{y^{3}}\). Wait, \(\sqrt{y^{3}}\cdot\sqrt{y^{3}}=(\sqrt{y^{3}})^{2}\). By the property \((\sqrt{a})^{2}=a\) for \(a\geq0\), so \((\sqrt{y^{3}})^{2}=y^{3}\)? Wait, no, that can't be. Wait, no, maybe the problem is \(\sqrt{y^{3}}\cdot\sqrt{y^{3}}\) is wrong. Wait, no, let's check the options. The options are \(2y^{3}\), \(y^{6}\), \(2y^{6}\), \(y^{3}\). Wait, maybe the problem is \(\sqrt{y^{3}}\cdot\sqrt{y^{3}}\) is actually \(\sqrt{y^{3}}\times\sqrt{y^{3}}\), but if we consider \(\sqrt{y^{3}}=y^{\frac{3}{2}}\), then \(y^{\frac{3}{2}}\times y^{\frac{3}{2}}=y^{\frac{3}{2}+\frac{3}{2}}=y^{3}\)? But that's not matching. Wait, maybe the problem is \(\sqrt{y^{3}}\cdot\sqrt{y^{3}}\) is a mistake, and it's \(\sqrt{y^{3}}\cdot\sqrt{y^{3}}\) should be \(\sqrt{y^{3}}\times\sqrt{y^{3}}\), but let's check the options again. Wait, the options include \(y^{3}\). Wait, let's do it step by step correctly.

Wait, \(\sqrt{y^{3}}=y^{\frac{3}{2}}\) (since \(\sqrt{a}=a^{\frac{1}{2}}\) and \((a^{m})^{n}=a^{mn}\), so \((y^{3})^{\frac{1}{2}}=y^{\frac{3}{2}}\)). Then \(\sqrt{y^{3}}\cdot\sqrt{y^{3}}=y^{\frac{3}{2}}\times y^{\frac{3}{2}}=y^{\frac{3}{2}+\frac{3}{2}}=y^{3}\). So the answer is \(y^{3}\) (the last option).

Wait, but let's check again. If \(y\geq0\), \(\sqrt{y^{3}}\cdot\sqrt{y^{3}}=\sqrt{y^{3}\times y^{3}}=\sqrt{y^{6}}\). And \(\sqrt{y^{6}}=y^{3}\) because \(y\geq0\) (if \(y\) could be negative, it would be \(|y^{3}|\), but since \(y\geq0\), \(|y^{3}| = y^{3}\)). So yes, the product is \(y^{3}\).

Answer:

The correct option is the last one (the option with \(y^{3}\))