Which is greater? A = (3)(2^{1/3}) or B = 1 + (2)(3^{1/3})?

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Markolainen

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The question is as follows:

If A [MATH]= 3 \cdot 2^{\dfrac{1}{3}}[/MATH], and B = [MATH]1 + 2 \cdot 3^{\dfrac{1}{3}}[/MATH]
Which is greater?

Answer:
A < B

Calculator is not accepted on the test.

Thanks in advance. :)
 
I would cube both sides, approximate 31/3 as 1+(more than 1 so for example 7*1+ would be more than 7) and approximate 32/3 = 91/3 as 2+.

Alternatively you can subtract 2*31/3 from both sides and factor out the gcd and .....

Please share your work with us.
 
Thank you for your reply. I tried your way but got it wrong. My approximation of [MATH]3^{1/3}[/MATH] being [MATH]1^{+}[/MATH] was more like 1.1 and not the 1.44 it really is. Since this test uses a lot of smaller numbers in n-th root maybe memorizing some of the basic ones could help.

But there is no "way" to solve these kind of problems mathematically?
 
I wish someone could tell me what the pedagogical point of this problem is.
The fact is, these two differ by some \(\displaystyle 0.1047\), without a calculator I fail to see any point to the question.
I do know that testing companies face huge problems with phones and even smart watches which are capable of supporting computer algebra systems, WolfFramAlpha for one.
 
pka said:
I wish someone could tell me what the pedagogical point of this problem is.
The fact is, these two differ by some \(\displaystyle 0.1047\), without a calculator I fail to see any point to the question.
I do know that testing companies face huge problems with phones and even smart watches which are capable of supporting computer algebra systems, WolfFramAlpha for one.
Click to expand...
The test itself is used as an alternative way of being eligible for some universities here in Sweden. Many of the questions are made in a way to challenge your understanding in how math works.
 
Markolainen said:
The test itself is used as an alternative way of being eligible for some universities here in Sweden. Many of the questions are made in a way to challenge your understanding in how math works.
Click to expand...
Thank you for that reply. I whole heartily support that aim. However, I cannot see how this particular question has any relation to that aim.
 
I think when I did it I got something like 3--2+. Now any number less than 3 minus any number more than 2 is always less than 1 which is exactly what you had to show. This worked out fairly easily. I am very surprised that my method worked when I read pka's reply saying how close the numbers were to one another.

Try the 2nd method of factoring.
 
Jomo said:
I think when I did it I got something like 3--2+. Now any number less than 3 minus any number more than 2 is always less than 1 which is exactly what you had to show. This worked out fairly easily. I am very surprised that my method worked when I read pka's reply saying how close the numbers were to one another. Try the 2nd method of factoring.
Click to expand...
You do realize that \(\displaystyle 3\sqrt[3]2<1+2\sqrt[3]3\)? I could not tell from the above. SEE HERE be sure to click \(\displaystyle \boxed{\text{approximate form }}\)
 
I note that calculators are not allowed in this thread.

My first thought was the same as Jomo's (i.e., compare A3 and B3) because we can evaluate A3 easily. It's 54.

When I saw Jomo's notation 1+, it made me consider using the binomial theorem for B3=(1+x)3 where x=2∙3⅓=2+, but then I wondered how to sufficiently approximate 3⅓ for a good estimate of 2+.

Clearly, 3⅓ must be between 1 and 2, so I evaluated 1.53 to start because I already knew 15^2=225.

153 = 225(15), and I did that multiplication mentally as 225(10+5)=2250+1125.

Therefore, 1.53 is 3.375 (too big).

Going through the same process with 1.4 but using paper and pencil for 196(10+4), I got 2744.

Therefore 1.4^3 is 2.744 (too small).

That all took less than two minutes. Next, I was about to use (1+x)=1+3x+3x2+x3 with x=2.8 when I realized it would be just as easy to multiply out (1+2.8)^3.

3.83 = 54.872

54.872 > 54 hence B > A

?
 
Using paper and writing instrument, I figured that

\(\displaystyle 1.2 < \sqrt[3]{2} < 1.3 \ \ \ and \ \ \ 1.4 < \sqrt[3]{3} < 1.5, \)

but using the upper bound of 1.3 and the lower bound of 1.4, respectively,
the answer to the question could not be resolved according to the manner I am showing:

3(1.3) vs. 1 + 2(1.4)
3.9 vs. 3.8

Using the other three cases, the left side is shown to be less than the right side at one decimal place.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

\(\displaystyle \ \ 3\sqrt[3]{2} \ \ \ vs. \ \ \ 1 + 2\sqrt[3]{3} \ \ \ \ \) *


So, let's look at cubing each side instead.

Let \(\displaystyle \ x = 2\sqrt[3]{3} \ \ \) in the expression\(\displaystyle \ \ (1 + x)^3\).

\(\displaystyle (1 + x)^3 = \)

\(\displaystyle 1 + 3x + 3x^2 + x^3 = \)

\(\displaystyle 3x + 3x^2 + 1 + x^3 = \)

\(\displaystyle 3x(1 + x) + (1 + x^3) \ \ \ \ \)**

Cubing the left-hand side of * equals 54.

Substitute \(\displaystyle \ x = 2\sqrt[3]{3}, \ \ and \ \ use \ \ the \ \ lower \ \ bound \ \ of \ \ 1.4 \ \ for \ \ \sqrt[3]{3} \ \ in \ \ \ \ \) **

\(\displaystyle 3(2\sqrt[3]{3})[1 + 2\sqrt[3]{3}] + [1 + (2\sqrt[3]{3})^3] = \)

\(\displaystyle 6\sqrt[3]{3}[1 + 2\sqrt[3]{3}] + 25 \approx \)

\(\displaystyle 6(1.4)[1 + 2(1.4)] + 25 = \)

\(\displaystyle (8.4)(3.8) + 25 = \)

\(\displaystyle 31.92 + 25 = \)

\(\displaystyle 56.92 \)

\(\displaystyle 54 < 56.92 \)

Therefore, \(\displaystyle \ \ 3\sqrt[3]{2} \ < \ 1 + 2\sqrt[3]{3} \).
 
Here is what I did and it was quick.
(1 + 2*31/3)3 = 1 + 3*2*31/3 + 3*22*91/3 + 24 >1 + 6 + 12*2 + 24 = 55. The lhs cubed is only 54.
I approximated 31/3 as just more than 1 and approximated 91/3 as simply more than 2.

There is no reason to get a better estimate for 31/3 and 91/3
I saw this because the easily calculable values on the right hand side was 25 and I saw the other terms were more than 29, which was enough to conclude that A<B
 
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