Answer:
[tex]\large\boxed{72+28\sqrt[3]3}[/tex]
Step-by-step explanation:
[tex]4\sqrt[3]3(7+6\sqrt[3]9)\qquad\text{use distributive property}\ a(b+c)=ab+ac\\\\=(4\sqrt[3]3)(7)+(4\sqrt[3]3)(6\sqrt[3]9)\qquad\text{use}\ \sqrt[3]{a}\cdot\sqrt[3]{b}=\sqrt[3]{ab}\\\\=(4\cdot7)(\sqrt[3]3)+(4\cdot6)(\sqrt[3]{(3)(9)})\\\\=28\sqrt[3]3+24\sqrt[3]{27}\qquad\sqrt[3]{27}=3\ \text{because}\ 3^3=27\\\\=28\sqrt[3]3+24(3)\\\\=28\sqrt[3]3+72[/tex]