Question

Math

Answer 1

\[\sqrt[3]{3xy^3} *\sqrt[3]{18x^5y^3}\]@hero

Answer 2

@hero

Answer 3

For this, it is best to look at the factors under the root to determine the exponent value. If the exponent value is less than 3, then that factor cannot be cube rooted. If the factor is 3 or greater, then it can be cube-rooted. However, there is a process to doing even that. What we will do is re-write the expressions according to this rule:
\(\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}\)

Answer 4

But we will do a bit of organizing here. We will let n = 3, and a represent the factors with exponents less than 3 and b will represent factors with exponents with a value of 3 or greater:

Answer 5

Your words confuse me but I am sure once I see the equation written out I will understand.

Answer 6

First though even before we do that, let's acknowledge the rule again above in this way:
\(\sqrt[n]{a}\sqrt[n]{b}=\sqrt[n]{ab}\)

Answer 7

So \[\sqrt[3]{3xy^318x^5y^3}\]

Answer 8

YES

Answer 9

Happy?

Answer 10

But now we will have to combine like terms next. So in the next step pair like terms together. \(x's\) with \(x's\), \(y's\) with \(y's\) numbers with numbers.

Answer 11

Because what you will do after that is apply the following rule:
\(a^ba^c = a^{b + c}\)

Answer 12

\[\sqrt[3]{21x^6y^6}\]

Answer 13

\(18 \times 3\) = \(21\)?

Answer 14

How'd you get that?

Answer 15

I was combining the terms under the radical so I thought I added, but no \[3*18=54\]

Answer 16

You only add exponents. \(3\) and \(18\) are not exponents of course.

Answer 17

\[\sqrt[3]{ 54x^6y^6}\]

Answer 18

@hero?

Answer 19

The only thing is now we will apply this rule in reverse now:
\(\sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b}\)

Also there is another rule to apply which is:
\(\sqrt[n]{a^b} = a^{b/n}\)

Answer 20

Sooo
\[\sqrt[3]{54}\]\[\sqrt[3]{x^6y^6}\]

Answer 21

Then \[54^2\] and \[x^2y^2\]

Answer 22

\[54^2x^2y^2\]

Answer 23

Well, the \(x^2\) and \(y^2\) is correct, but \(54^2\) is not because actually I did not intend for you to jump that far ahead.

Answer 24

Sorry..

Answer 25

\(\sqrt[3]{54}\) and \(\sqrt[3]{x^6y^6}\) So as you understood before, according to the rule we can rewrite \(\sqrt[3]{x^6y^6}\) as \(x^{6/3}y^{6/3}\) which comes to just \(x^2y^2\), but notice that there is no exponent for \(54\) so we have to do something else with it.

Answer 26

Find the cubed root?

Answer 27

Yes, but it is best to first put it in some form that will easily allow us to take the cube root. Right now 54 is not in a correct form to cube root.

Answer 28

What we have to do is do that factoring process I showed you earlier for factoring quadratic polynomials. We keep factoring 54 starting with 1 and 54 until we find a factor of 54 that is a cube as follows:

(1)(54) (no cubes here)
(2)(27) (27 is a cube so next step write it in cube form:)
\((2)(3^3)\)

Answer 29

So that means:
\(\sqrt[3]{54} = \sqrt[3]{(2)(3^3)}\)

Answer 30

Which we can use the rule \(\sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b}\) to further simplify it as follows:
\(\sqrt[3]{(2)(3^3)} = \sqrt[3]{2}\sqrt[3]{3^3}\)

Answer 31

Then use the rule I used on the exponents?

Answer 32

And then we can use the rule \(\sqrt[n]{a^n} = a\) to simplify further

Answer 33

So you end up with \(3\sqrt[3]{2}x^2y^2\) as the complete cube root of the original expression.

Answer 34

Woah