Operations with cube roots, fourth roots, and other higher-index roots work similarly to square roots, though, in some spots, we'll need to extend our thinking a bit. I'll explain as we go.
In the previous pages, we simplified square roots by taking out of the radical any factor which occurred in sets of two. For the second root, we needed a second copy.
For higher-index roots, the thinking is the same. If we have a cube root, we can take out any factor that occurs in sets of three; in a fourth root, we take out any factor that occurs in sets of four; in a fifth root, we take out any factor that occurs in sets of five; and so forth. For instance:
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Previously, I could pull from a square root anything that of which I had two copies. In the same way now, I can pull from a fourth root anything of which I've got four copies. Since 16 = 2^{4}, then:
I'm taking the cube root. I can then pull out of the radical any factor that occurs three times. Since 8 = 2^{3}, then this radical will simplify completely.
My first step is to factor this completely:
54 = 2 · 27 = 2 · (3 · 3 · 3)
I have three copies of 3, so I can pull a 3 out of the cube root, leaving a 2 inside.
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Again, I start with factoring:
48 = 3 · 16 = 3 · 2 · 2 · 2 · 2
I have four copies of the factor 2, but this is a cube root, so I can only pull out a 2 for three of these copies. The 3 and the fourth 2 will remain behind inside the radical.
I know that 27 = 3^{3}, so the cube root will simplify to a whole number. Then I'll finish by multiplying.
They've given me variables inside this radical, but the process works the same way as always. I'm taking a fifth root, so I can pull out of the radical anything for which I've got five copies.
The 32 is 2^{5}, so this will come out of the radical. The x^{10} = (x^{2})^{5}, so an x^{2} will come out. The y^{6} = (y^{5})(y^{1}), so I'll be able to pull a y out, while leaving the last y inside the radical. And z^{7} = (z^{5})(z^{2}), so I'll be able to pull a z out, while leaving a z^{2} inside.
My work looks like this:
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Note: When you're simplifying radical expressions with variables, if the radical is an even-index root (like a square root or a fourth root), they'll probably specify that you should "assume that all variables are non-negative" (or "positive"). This is to avoid having to take into consideration whether or note absolute-value bars are necessary on your answer. If you aren't sure what I'm talking about, check here.
This multiplication works just like the multiplication of square roots, in that the product of two of the same higher-index root can be converted to the higher-index root of the product. Then I simplify in the usual manner.
In this case, they've given me a product of fourth roots. I can convert the product of radicals into the radical of a product. Then I can simplify.
The terms in this expression are both cube roots, but I can combine them only if they're the cube roots of the same value. Right now, they aren't. So I'll simplify the radicals first, and then see if I can go any further.
I note that 8 = 2^{3} and 64 = 4^{3}, so I will actually be able to simplify the radicals completely.
I can't simplify the second radical at all. But I can simplify the first radical, because 81 = 3^{4} = (3^{3})(3). So I will end up with the sum of two third roots of three, which I can combine.
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The denominator is a cube, being 27 = 3^{3}, so I can easily simplify and arrive at a "rationalized" denominator:
This is similar to the previous exercise, but here the cube (that is, the 27) is in the numerator. I can't simplify this expression properly, because I can't simplify the radical in the denominator down to whole numbers:
To rationalize a denominator containing a square root, I needed two copies of whatever factors were inside the radical. For a cube root, I'll need three copies. So that's what I'll multiply onto this fraction.
I have one copy of the factor 5 in the denominator. I'll multiply, top and bottom, by the cube root of 25, which will provide the additional two copies of 5 that I need, in order to rationalize the denominator.
This final expression is arguably not much "simpler" than the original expression. In this context, "simplify" meant "rationlize the denominator". Oftentimes, the "right" answer won't be much, if at all, "simpler" than what you'd started with.
Since 72 = 8 × 9 = (2 × 2 × 2) × (3 × 3), I have only three 2's and two 3's. In other words, as the fraction currently stands, I won't have enough of any of the denominator's factors to get rid of the radical.
To take anything out of a fourth root, I would need four copies of each factor. For this denominator's radical, I'll need two more 3's and one more 2. So I'll multiply, top and bottom, by the fourth root of 3 · 3 · 2 = 18.
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