Question

simplify 216 2/3

Answer 1

Just divide it lol lo l ol o l

Answer 2

\[216^{2/3}\]Raising to that power is squaring, then taking the cube root, or alternatively, taking the cube root, then squaring.

\[216^{2/3} = \sqrt[3]{216^2}\]
Factor 216 into its prime factors and you should be able to see how to proceed.

Answer 3

No need for the lol. I haven't been in school for over 15 years and I am trying to help my daughter.

Answer 4

I would personally try to do the cube root first.

Answer 5

I don't understand I am sorry.

Answer 6

Okay, do you know what a square root is?

Answer 7

Yes.

Answer 8

How about a cube root?

Answer 9

no I don't know what a cube root is

Answer 10

I just googled it...so yes I do know now

Answer 11

A cube root is like a square root, except instead of there being 2, there are 3 factors. So the cube root of 216 will be the number x such that x*x*x = 216.

Answer 12

One way to find it is to decompose x into all of its factors.

216 = 2 * 108 = 2*2*54 = 2*2*2*27 = 2*2*2*3*9 = 2*2*2*3*3*3

With a cube root, each trio of 3 identical factors under the cube root sign turns into 1 of the identical factor outside the cube root sign. For example

\[\sqrt[3]{8} = \sqrt[3]{2*2*2} = 2\]

Can you find the cube root of 216?

Answer 13

6

Answer 14

It is multiple choice. When it is multiple choice it is easier to find the answer, but I wouldn't have a clue if they didn't give answers to choose from.

Answer 15

its 36

Answer 16

not 6

Answer 17

Yes! I was about to say that you fell into the trap :-)

Answer 18

So it's not 6x6x6 its 36x3x2....how confusing. They did not have this when I was in highschool

Answer 19

So if there is one that has you adding 2 of those expressions would you just add them together? What about 64 1/3 + 81 1/2. Let me see if I understand this.

Answer 20

Now, for a learning exercise, why don't we see if the order matters? We took the cube root first, and then squared it. Let's try a different problem where the numbers are a little smaller, and this time we will square it first, then take the cube root.

what's 36*3*2? the factorization of 216? I'll come back to that.

How about 27? We'll do \[\sqrt[3]{27^2}\] but we'll square 27 first, then take the cube root. 27 has factors \(3*3*3\), and we are squaring it, so we'll write the factors twice:
\[\sqrt[3]{27^2} = \sqrt[3]{(3*3*3)^2} = \sqrt[3]{(3*3*3)*(3*3*3)}\]
With me so far?

Answer 21

Yes

Answer 22

Okay, now we are going to take a cube root, so we look for trios of equal factors. Well, pretty easy to see them the way I wrote it, there are 2 groups of 3 3's, so our cube root is just 3*3 = 9.

27^2 = 729, and 9^3 = 729. We can do the operations in either order. Does that make sense?

Answer 23

The bottom part does not make sense. What is that little arrow thing for?

Answer 24

You asked about \[64^{1/3}+81^{1/2}\]which we can also write with radical signs as \[\sqrt[3]{64} + \sqrt{81}\]Again, we factor each number into its prime factors(*)
\[\sqrt[3]{(2*2*2)*(2*2*2)} + \sqrt{(3*3)*(3*3)} = 2*2 + 3*3 = 13\]

Answer 25

Sorry, I was a bit lazy, the 27^2 is just \(27^2\) without all the formatting characters I have to type to make it look pretty :-)

Answer 26

Similarly 9^3 is \(9^3\)

Clear as mud?

Answer 27

Is there a website I can go to and read on how to do this?

Answer 28

Oh, probably. But do you have a public library nearby? There's a DVD my son just watched by a group called the Standard Deviants, the title of the DVD is Pre-Algebra part 1, and they covered this exact material.

Answer 29

I have my daughter's textbook, but they don't really show anything.

Answer 30

I could also just go through it with you right now. I taught it to my 9-year-old, I'm sure I can teach it to you :-)

Answer 31

If you wouldn't mind.

Answer 32

If a 9 year old gets it then a 31 year old has to right?

Answer 33

Of course!

Let me outline the basics of exponents.

If you have an expression \(a^b\), \(a\) is the base number, and \(b\) is the exponent. It's a shorthand of sorts: with integer exponents, you can think of \(a^b\) as \(b\) \(a\)'s all lined up in a row with multiplication signs between them. As we saw, \(2^3 = 2*2*2\), for example.

Any non-zero number raised to the 0 power = 1. \(a^0 = 1, a\ne0\).

Any number raised to the 1 power = itself. \(a^1 = 1\)

A number raised to a negative power can be rewritten as the reciprocal of the number raised to the positive power. \(a^{-n} = \dfrac{1}{a^n}\)

Similarly, you can rewrite a negative exponent in the denominator as a positive exponent in the numerator. \(\dfrac{1}{a^{-n}} = \dfrac{a^{n}}{1} = a^n\)

If you multiply two exponential expressions which have the same base number, you keep the base number and add the exponents. \(a^n*a^m = a^{n+m}\)

If you divide two exponential expressions which have the same base number, you keep the base number and subtract the exponents (numerator - denominator). \(\dfrac{a^n}{a^m} = a^{n-m}\)

If you multiply two exponential expressions which have the same exponent but different base numbers, you keep the exponent and multiply the base numbers. \(a^n*b^n = (ab)^n\)

Answer 34

If you divide two exponential expressions which have the same exponent but different base numbers, you keep the exponent and divide the base numbers (numerator/denominator). \(\dfrac{a^n}{b^n} = (\dfrac{a}{b})^n\)

Answer 35

@christina0603 have I put you into a sound slumber yet? :-)

Answer 36

Would you like some examples of all of these rules in action?

Answer 37

I am here.

Answer 38

Yes a few examples would help.

Answer 39

Okay, I need a few minutes to fix a snack to sustain me through these Herculean efforts :-)

Answer 40

I think I will need one too.

Answer 41

I'd like to take this opportunity to say a word of thanks for the engineer who put the "Potato" button on my microwave oven :-)

Answer 42

I know I love it.

Answer 43

Okay, on to the examples!

I trust \(a^0 = 1\) and \(a^1 = a\) are self-explanatory. If you don't agree, I'm at a loss to provide an example that would make it otherwise :-)

Number raised to a negative power:

\[2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4}\]I hear you thinking, "but how do we know that?" I need to give examples of a few more things before I show that.

Multiplying 2 expressions with the same base number:
\[2^3*2^4 = (2*2*2)*(2*2*2*2) = 2^7 = 2^{3+4}\]

Dividing 2 expressions with the same base number:
\[\frac{2^5}{2^3} = \frac{2*2*2*2*2}{2*2*2} = 2*2 = 4 = 2^2 = 2^{5-3}\]

Let's divide two numbers with the denominator a bigger exponent than the numerator:
\[\frac{2^2}{2^4} = \frac{2*2}{2*2*2*2} = \frac{1}{2*2} = \frac{1}{4}\]But wait, what about our exponent subtraction? Well, \[\frac{2^2}{2^4} = 2^{2-4} = 2^{-2} = \frac{1}{4} = \frac{1}{2^2}\]If you're still not convinced, consider what happens if we take \(2^1\) and divide it by 2, 4, 8, etc.
\[\frac{2^1}{2} = \frac{2^1}{2^1} = \frac{2}{2} = 2^{1-1} = 2^0 = 1\]\[\frac{2^1}{4} = \frac{2^1}{2^2} = \frac{2}{4} = 2^{1-2} = 2^{-1} = \frac{1}{2}\]\[\frac{2^1}{8} = \frac{2^1}{2^3} = \frac{2}{8} = 2^{1-3} = 2^{-2} = \frac{1}{4}\]etc. Hopefully that convinces you about converting between negative and positive exponents.

Answer 44

Sorry, I've been experiencing some browser issues :-(

Answer 45

Multiplying exponential expressions with different base numbers but same exponent:
\[2^3*3^3 = 8*27 = 216 = (2*2*2)*(3*3*3) = (2*3)*(2*3)*(2*3) = \]\[(2*3)^3 = 6^3 = 6*6*6 = 216\]

Dividing exponential expressions with different base numbers but same exponent:
\[\frac{4^3}{2^3} = \frac{4*4*4}{2*2*2} = \frac{64}{8} = \frac{4}{2}*\frac{4}{2}*\frac{4}{2} = (\frac{4}{2})^3 = 2^3 = 8\]

Answer 46

Oops, I forgot an important law:

\[(a^n)^m = a^{n*m}\]\[(2^3)^4 = (2*2*2)^4 = (2*2*2)*(2*2*2)*(2*2*2)*(2*2*2) = 2^{12} = 2^{3*4}\]

Answer 47

It's getting a bit late, and I haven't heard anything from you in a while, so I'll just do a brief bit on square roots.

Remember that the square root of x is just a number which multiplied by itself gives us x. \[\sqrt{x} * \sqrt{x} = x\]From that, we can see that \(\sqrt {(x^2)} = x\), because \(x*x = x^2\). Now let's think back to the bit about multiplying numbers with the same base number. \[x = x^1\]\[x^2 = x^1*x^1 = x^{1+1}\]\[\sqrt{x^2} = \sqrt{x^1*x^1} = x^1\]\[x^3 = x^1*x^1*x^1\]\[(x^3)^2 = x^1*x^1*x^1*x^1*x^1*x^1 = x^6\]\[\sqrt{x^6} = \sqrt{x^3*x^3} = x^3\]See the pattern? Taking the square root of some exponential expression is just dividing the exponent by 2. If the exponent was already 1, taking the square root means\[\sqrt{x} = \sqrt{x^1} = x^{1/2}\]By similar means, we can see that a cube root is \[\sqrt[3]{x^3} = \sqrt[3]{x*x*x} = x = x^1 = x^{3/3}\] so we divide the exponent by 3 to take a cube root. \[\sqrt[3]{x} = x^{1/3}\]\[(x^{1/3})^3 = x^{3*(1/3)} = x^{(1/3)*3} = x^1 = x\]
Cubing the cube root gives us x, just as it should.

Answer 48

Simplifying square roots:

Sometimes when you factor what is under the radical sign, you won't have a product of 1 or more perfect squares. Take for example \[\sqrt{27} = \sqrt{3*3*3}\]We can factor out 1 pair of 3's, but we still have a lonely 3 trapped under the radical sign. We have to leave it there:\[\sqrt{27} = \sqrt{3*3*3} = \sqrt{(3*3)*3} = 3*\sqrt{3} \approx 3*1.73205 \approx 5.19615 \]and if we punch 5.19615*5.19615 into our trusty calculator, we get 26.999974823 which is pretty close to 27, I think you'll agree. Unless the answer calls for a number, or you are in a situation where a number would be expected (a physics or chemistry problem, for example), I would leave it in the symbolic form \(3\sqrt{3}\).

If you have a fraction with a square root in the denominator, such as \(\dfrac{1}{\sqrt{x}}\), that can be written as \[\frac{1}{\sqrt{x}}*\frac{\sqrt{x}}{\sqrt{x}} = \frac{\sqrt{x}}{x}\]

Sometimes you'll get an expression like \[\frac{1}{1-\sqrt{x}}\]Here you would multiply by the conjugate of the denominator:
\[\frac{1}{1-\sqrt{x}}*\frac{1+\sqrt{x}}{1+\sqrt{x}} = \frac{1 + \sqrt{x}}{1^2 + \sqrt{x} - \sqrt{x} + \sqrt{x}\sqrt{x}} = \frac{1+\sqrt{x}}{1+x}\]

Answer 49

Print that out, put it under your pillow, and in the morning you will once again be a whiz at this stuff :-)