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pepetheskunkcat
The answer is [tex]13 \sqrt[4]{x} -10 \sqrt[4]{2y} [/tex]
ThonyBossa

Answer:

[tex]13\sqrt[4]x - 10\sqrt[4]2y[/tex]

Step-by-step explanation:

[tex]2(\sqrt[4]16x) - 2(\sqrt[4]2y) + 3(\sqrt[4]81x) - 4(\sqrt[4]32y)[/tex]

In this case, we need to get the Least Common Multiple (LCM) of 16, 81 and 32, that is:

  • [tex]16 = 2^4[/tex]
  • [tex]81 = 3^4[/tex]
  • [tex]32 = 2^5[/tex]

Now, we replace in the original equation, that is:

[tex]2\sqrt[4](2^4x) - 2\sqrt[4](2y) + 3\sqrt[4](3^4x) - 4\sqrt[4](2^5y)[/tex]

Simplifying

[tex]4\sqrt[4]x + 9\sqrt[4]x - 2\sqrt[4]2y - 8\sqrt[4]2y[/tex]

Finally, we have

[tex]13\sqrt[4]x - 10\sqrt[4]2y[/tex]

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