SOLUTION: Multiply ^3(over the square root of)y^10 by ^3(over the square root of)16y^11 and simplify by factoring while assuming that all expressions under radicals represent nonnegative num
Algebra ->
Square-cubic-other-roots
-> SOLUTION: Multiply ^3(over the square root of)y^10 by ^3(over the square root of)16y^11 and simplify by factoring while assuming that all expressions under radicals represent nonnegative num
Log On
Question 401458: Multiply ^3(over the square root of)y^10 by ^3(over the square root of)16y^11 and simplify by factoring while assuming that all expressions under radicals represent nonnegative numbers. Answer by jsmallt9(3758) (Show Source):
What you are calling a square root is actually the radical symbol. The radicals symbol is used for all types of roots. The little number in the upper left corner of the radical is called the index. The index indicates which type of root the radical represents. The index of 3 on your radicals tells us that they represent cube (or 3rd) roots. A radical without a visible index has an implied index of 2 and represents a square (or 2nd) root. This is why radicals and square roots can be easily confused.
When multiplying radicals you have to choice of
Simplify, multiply and simplify again; or
Multiply and simplify
The first approach seems like more work. But when the multiplication is complicated it can help a lot if you have simplified first. Your multiplication is not very complicated so we will use the second approach.
Multiplying radicals uses a basic property of radicals: . Using this to multiply your radicals we get:
which simplifies to
We can simplify this further if we can find perfect cube factors in the radicand (the expression within a radical). For this means
looking for perfect cube factors of 16; and/or
finding perfect cube factors of . Because of the way exponents work, any exponent that is a multiple of 3 is a perfect cube.
8 is a perfect cube (because it is ) and 8 is a factor of 16. is a perfect cube itself not because 21 is a perfect cube (which it is not) but because 21 is a multiple of 3. So we can factor the radicand as follows:
For reasons you will see shortly, I like to use the Commutative Property to rearrange the order of the factors so that the perfect cube factors are in front:
Now we use the same property as earlier. But this time we are using it in the opposite direction to take a single cube root of a product and break it into a product of the cube roots of each factor:
The cube roots of the perfect cubes simplify:
Note how the radical is at the end. This is the normal way to write terms like this. The radical is at the end now because we put the perfect cube factors in front earlier.
Note: The part about "non-negative numbers" was irrelevant in this problem. The stipulation of "non-negative numbers" has meaning only for even-numbered roots (square roots, 4th roots, 6th roots, etc.). Cube (or 3rd) roots are odd-numbered roots and negative vs. non-negative is not an issue for odd-numbered roots.