9514 1404 393
Answer:
Step-by-step explanation:
We can call the blank values 'a' and 'b' so we have ...
[tex]ay\sqrt[3]{6y}-14\sqrt[3]{48y^b}=-11y\sqrt[3]{6y}[/tex]
We can move everything to inside the radical by cubing those things that are outside.
[tex]\sqrt[3]{6a^3y^4}-\sqrt[3]{131712y^b}=-\sqrt[3]{7986y^4}[/tex]
In order for the terms on the left to be "like" terms, the value of b must be 4. Having determined that, we can remove the cubes from under the radical and factor out the radical.
[tex]ay\sqrt[3]{6y}-28y\sqrt[3]{6y}=-11y\sqrt[3]{6y}\\\\a-28=-11 \qquad\text{divide by $y\sqrt[3]{6y}$}\\\\a=17[/tex]
Then the desired expression is ...
[tex]\boxed{17}y\sqrt[3]{6y}-14\sqrt[3]{48y^{\boxed{4}}}=-11y\sqrt[3]{6y}[/tex]
