1. State the problem: Simplify $\sqrt[3]{\frac{49}{27}}\, j^{1/3}$. 2. Use the cube root rule: $\sqrt[3]{\frac{49}{27}}\, j^{1/3}=\sqrt[3]{\frac{49}{27}}\,\sqrt[3]{j}$ 3. Simplify $\sqrt[3]{\frac{49}{27}}$ by separating the fraction: $\sqrt[3]{\frac{49}{27}}=\sqrt[3]{\frac{49}{1}}\,\sqrt[3]{\frac{1}{27}}=\sqrt[3]{49}\,\sqrt[3]{\frac{1}{27}}$ 4. Rewrite $49$ and $27$ as cubes times leftovers: $49=7\cdot 7\cdot 1\quad\text{and}\quad 27=3\cdot 3\cdot 3$ 5. Evaluate the cube roots: $\sqrt[3]{49}=\sqrt[3]{7^2}=\sqrt[3]{7^2}$ $\sqrt[3]{\frac{1}{27}}=\frac{\sqrt[3]{1}}{\sqrt[3]{27}}=\frac{1}{3}$ 6. Combine the real part: $\sqrt[3]{\frac{49}{27}}=\sqrt[3]{49}\cdot\frac{1}{3}=\frac{\sqrt[3]{49}}{3}$ 7. Simplify the full expression: $\sqrt[3]{\frac{49}{27}}\, j^{1/3}=\left(\frac{\sqrt[3]{49}}{3}\right)\sqrt[3]{j}$ 8. Final answer: $\frac{\sqrt[3]{49}}{3}\sqrt[3]{j}$