MATH SOLVE

2 months ago

Q:
# What is the following sum?(please show how you worked it out)

Accepted Solution

A:

Answer:[tex]4\sqrt[3]{2}x(\sqrt[3]{y}+3xy\sqrt[3]{y} )[/tex]Step-by-step explanation:Let's start by breaking down each of the radicals:[tex]\sqrt[3]{16x^3y}[/tex]Since we're dealing with a cube root, we'd like to pull as many perfect cubes out of the terms inside the radical as we can. We already have one obvious cube in the form of [tex]x^3[/tex], and we can break 16 into the product 8 Β· 2. Since 8 is a cube root -- 2Β³, to be specific, we can reduce it down as we simplify the expression. Here our our steps then:[tex]\sqrt[3]{16x^3y}\\=\sqrt[3]{2\cdot8\cdot x^3\cdot y}\\=\sqrt[3]{2} \sqrt[3]{8} \sqrt[3]{x^3} \sqrt[3]{y} \\=\sqrt[3]{2} \cdot2x\cdot \sqrt[3]{y} \\=2x\sqrt[3]{2}\sqrt[3]{y}[/tex]We can apply this same technique of "extracting cubes" to the second term:[tex]\sqrt[3]{54x^6y^5} \\=\sqrt[3]{2\cdot27\cdot (x^2)^3\cdot y^3\cdot y^2} \\=\sqrt[3]{2}\sqrt[3]{27} \sqrt[3]{(x^2)^3} \sqrt[3]{y^3} \sqrt[3]{y^2}\\=\sqrt[3]{2}\cdot 3\cdot x^2\cdot y \cdot \sqrt[3]{y^2} \\=3x^2y\sqrt[3]{2} \sqrt[3]{y}[/tex]Replacing those two expressions in the parentheses leaves us with this monster:[tex]2(2x\sqrt[3]{2}\sqrt[3]{y})+4(3x^2y\sqrt[3]{2} \sqrt[3]{y})[/tex]What can we do with this? It seems the only sensible thing is to look for terms to factor out, so let's do that. Both terms have the following factors in common:[tex]4, \sqrt[3]{2} , x[/tex]We can factor those out to give us a final, simplified expression:[tex]4\sqrt[3]{2}x(\sqrt[3]{y}+3xy\sqrt[3]{y} )[/tex]Not that this is the same sum as we had at the beginning; we've just extracted all of the cube roots that we could in order to rewrite it in a slightly cleaner form.