How do you subtract \root[ 3] { 32x ^ { 3} y ^ { 4} } - 3x y \root [ 3] { 4y }?

Mar 19, 2018

Pull as much as you can outside of the first radical, and then group to find your answer, $- x y \setminus \sqrt[3]{4 y}$

Explanation:

The first thing to do is simplifying the left-hand set of terms. I treat multiplication radicals as their own sets for simplicity, and then I recombine afterwards:

$\setminus \sqrt[3]{32 {x}^{3} {y}^{4}} = \setminus \sqrt[3]{32} \times \setminus \sqrt[3]{{x}^{3}} \times \setminus \sqrt[3]{{y}^{4}}$

$\Rightarrow 2 \setminus \sqrt[3]{4} \times x \times y \setminus \sqrt[3]{y} = 2 x y \setminus \sqrt[3]{4 y}$

Now our equation looks like this:

$2 x y \setminus \sqrt[3]{4 y} - 3 x y \setminus \sqrt[3]{4 y}$

Next, we'll group the like factor $x y \setminus \sqrt[3]{4 y}$:

$2 x y \setminus \sqrt[3]{4 y} - 3 x y \setminus \sqrt[3]{4 y} = \left(x y \setminus \sqrt[3]{4 y}\right) \left(2 - 3\right)$

Finally, we'll do the simple arithmetic in the right-hand parentheses and we'll be done!

2-3=-1 rArr (-1)(xy\root[3]{4y}) rArr color(red)(-xy\root[3]{4y}