Simplify the Cube Root of 27/16 =? Algebra Problem - MOST will NOT Get RIGHT!

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Excellent explanation, very helpful. Highly appreciated

abdulqadersaidi
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(27/16)^1/3

27^1/3 = 3
16=2x2x2x2
(2x2x2x2)^1/3=
2x2^1/3

3 divided by 2x2^1/3

3x(2x2)^1/3 divided by 2x(2)^1/3x(2x2)^1/3

3x4^1/3 divided by
2x2

3 x 4^1/3 divided by 4

chrisdissanayake
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Great. Not new to math but forgot everything and want to understand what Roger Penrose is talking about which brought me here to get the basic sense of math.

BabarizamDK
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Way over my head but you explained it well!

gayschaye
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Thank you for the refresher on this! 😊

nancyholcombe
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Thank you. I understood surprisingly 😊

annathomas
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3/4 cube root 4. It is not difficult if you know that cube root 27/16 = cube root 27 : cube root 16. And don't forget about the fancy 1 to get rid of the (cube) root in the denominator.

Kleermaker
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Playing around, I think I see an interesting pattern for rationalizing denominators - I think this holds.

For instance, rationalize 13/(7th root of 18).

One answer is (13 * (7th root of (18^6))) / 18.

a/the b root of c = a * (the b root of c^(b-1)), all of that over c.

Of course, further simplification may be necessary.

I foresee an evening playing with logs.

johnnyragadoo
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Is this an Algebra Word Problem?? I ask that because it is in the playlist AWP. But it doesn't belong there.

Kleermaker
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great simplification thanks for the fun

russelllomando
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I haven't done algebra in over 40 years and this was easy.

tomhiggins
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[27/16]^(1/3)
27=3×3×3
16=2×2×2×2
[27]^(1/3)=3
[16]^(1/3)=2×(2)^(1/3)
to eliminate a radical in denominator requires multiplying both numerator and denominator by [(2)^(2/3)]
So
Numerator :
=(3)×[(2)^(2/3)]
Denomonator :
=(2)×[(2)^(1/3)][(2)^(2/3)]
=(2)×[(2)^(3/3)]
=4

Collected
[3×[(2)^(2/3)]]/[4]
Reformatted
[3×(third_rt(4))]/4

tomtke
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super simple lets start with numerator
(27) ^ 1/3 is the same as 3 * 3 * 3.
so numerator is 3 (1 ^ 1/3)

denominator:
(16) ^ 1/3 which is NOT a perfect cube but 8 as in 8 * 2 is a perfect cube
2(2 ^ 1/3)

Now we have a root in the denominator which is bad so we get rid of it by multiplying top and bottom by (2 ^ 1/3)
[ 3 ( 1 ^ 1/3 ) ( 2 ^ 1/3 ) ] / [ 2 ( 2 ^ 1/3 ) ( 2 ^ 1/3 ) ]
( 3 ( 2 ^ 1/3 ) / ( 2 * 2)

Therefore the answer is
( 3 ( 2 ^ 1/3 ) / 4

bigdog
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Interesting - another pattern.

The original problem is 3/16^(1/3), which is the same as 3 * 16^(-1/3).

I wish I understood this stuff a little better.

johnnyragadoo
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I saw that cubrt 27 =3
cubrt 16 = cubrt 4^2
To make 4^2 into 4^3 requires another 4
So
3 * cubrt 4= 3 cubrt 4
over
(cubrt 4^2) * (cubrt 4^1) = cubrt 4^3 = 4

vespa
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Before watching:

Well, 16=2^4, and (a^b)^c = a^(bc)

Thus, (27/16)^(1/3) = 3/(2^(4/3))

Psykolord
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³√(27/16)
= 3 / (2 ⋅ ³√2)
= 3 ⋅ ³√2² / 4
= 3 ⋅ ³√4 / 4

Nikioko
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Why not multiply both the numerator and denominator by 4.
One step
Answer will be straight forward

AliRahil-yqqb
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³√(27/16) no tiene solución en los números racionales, y cualquier intento de expresar en formato decimal siempre será una aproximación, nunca el resultado perfecto

lucasveren
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Demasiado enredado para algo tan sencillo.

marioguercio