Answer:
Option: A is the correct answer.
The sum is:
[tex]4x(\sqrt[3]{2y})+12x^2y(\sqrt[3]{2y^2})[/tex]
Step-by-step explanation:
We are given a expression as:
[tex]2(\sqrt[3]{16x^3y})+4(\sqrt[3]{54x^6y^5})[/tex]
Since, the quantity which comes three times inside the cube root sign comes out as a single.
( i.e. if we have:
[tex]\sqrt[3]{x^3}=x[/tex] )
Also,
[tex]2(\sqrt[3]{16x^3y})=2(2x\sqrt[3]{2y})\\\\i.e.\\\\2(\sqrt[3]{16x^3y})=4x(\sqrt[3]{2y})[/tex]
( Since,
[tex]16x^3y=2^3\cdot x^3\cdot 2y[/tex] )
Also,
[tex]4(\sqrt[3]{54x^6y^5})=4(3x^2y\sqrt[3]{2y^2})\\\\i.e.\\\\4(\sqrt[3]{54x^6y^5})=12x^2y(\sqrt[3]{2y^2})[/tex]
( Since,
[tex]54x^6y^5=3^3\cdot (x^2)^3\cdot y^3\cdot 2\cdot y^2[/tex] )
Hence, we get the simplified expression as:
[tex]4x(\sqrt[3]{16x^3y})+12x^2y(\sqrt[3]{2y^2})=4x(\sqrt[3]{2y})+12x^2y(\sqrt[3]{2y^2})[/tex]
