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Sedbober [7]
3 years ago
10

Someone please help me (asap)...

Mathematics
1 answer:
Dvinal [7]3 years ago
6 0

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}

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