Question

Someone please help me (asap)...

Answer 1

9514 1404 393

Answer:

Step-by-step explanation:

We can call the blank values 'a' and 'b' so we have ...

  $ay\sqrt[3]{6y}-14\sqrt[3]{48y^b}=-11y\sqrt[3]{6y}$

We can move everything to inside the radical by cubing those things that are outside.

  $\sqrt[3]{6a^3y^4}-\sqrt[3]{131712y^b}=-\sqrt[3]{7986y^4}$

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.

  $ay\sqrt[3]{6y}-28y\sqrt[3]{6y}=-11y\sqrt[3]{6y}\\\\a-28=-11 \qquad\text{divide by y\sqrt[3]{6y} }\\\\a=17$

Then the desired expression is ...

  $\boxed{17}y\sqrt[3]{6y}-14\sqrt[3]{48y^{\boxed{4}}}=-11y\sqrt[3]{6y}$