Question

1. which of the following is equivalent to $sqrt3{4k^2} cdot sqrt3{6jk}$?

• $2ksqrt{6kj}$
• $2ksqrt3{3j}$
• $2ksqrt3{3jk}$
• $2ksqrt3{6kj}$

Explanation:

Step1: Multiply the cube roots

Recall the property of radicals: \(\sqrt[3]{a}\cdot\sqrt[3]{b}=\sqrt[3]{ab}\). So, \(\sqrt[3]{4k^{2}}\cdot\sqrt[3]{6jk}=\sqrt[3]{4k^{2}\cdot6jk}\).

Step2: Simplify the product inside the cube root

Calculate \(4\times6 = 24\) and \(k^{2}\times k=k^{3}\), so the expression inside the cube root becomes \(24jk^{3}\). So we have \(\sqrt[3]{24jk^{3}}\).

Step3: Factor out the perfect cube

We know that \(24 = 8\times3\) and \(k^{3}\) is a perfect cube. So, \(\sqrt[3]{24jk^{3}}=\sqrt[3]{8k^{3}\cdot3j}\).

Step4: Simplify the cube root of the perfect cube

Using the property \(\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}\) again, \(\sqrt[3]{8k^{3}\cdot3j}=\sqrt[3]{8k^{3}}\cdot\sqrt[3]{3j}\). Since \(\sqrt[3]{8k^{3}} = 2k\), the expression simplifies to \(2k\sqrt[3]{3jk}\).

Answer:

\(2k\sqrt[3]{3jk}\) (The third option: \(2k\sqrt[3]{3jk}\))