De-nesting An Interesting Radical

[@matthewfeig5624]

You basically have the answer on the screen at 1:45 from your warm-up problem! cbrt( 7 + 5sqrt(2) ) = 1 + sqrt(2) Just multiply both sides by 2^(1/6) = cbrt( sqrt(2) ) to get the expression we want on the left. cbrt( 10 + 7sqrt(2) ) = 2^(1/6)*[ 1 + sqrt(2) ]. (This is the form of the answer that @Blaqjaqshellaq stated.)

[@stevenngov6405]

YOO THAT AWESOMEE

[@davidseed2939]

5:30 k=1/2 a^3=4. BUT HARD to.see. might be easier trying a=rb, results in r=2

[@dannkod]

Right. Or you can divide the polynomial 20k³ - 42k² + 30k - 7 = 0 with k³ and substitute y=1/k and have y=2 and then k=1/2 if you did b=ak substitution.

[@alipourzand6499]

Unexpected result! Hope once you will make a video about rationalizing cube roots in denominator.

[@lhdill2009]

Syber, I'm trying to understand the implications of your solution to related de-nesting problems. From what I understand, the norm of 10+7 sqrt(2) is (10+7 sqrt(2) )*(10-7 sqrt(2) ) = 100 - 98 = 2. Because the cube root of 2 is irrational, any attempt to write [10+7 sqrt(2) ]^(1/3) = a + b sqrt(2) will lead to irrational values of a and b rather than rational values, which we greatly prefer. This is because [10-7 sqrt(2) ]^(1/3) must be a - b sqrt(2) -- is this correct? -- so that the norm of these cube roots will be a^2 - 2 b^2 = 2^(1/3). Since the RHS of this expression is irrational, a or b or both a and b must also be irrational. Is this the gist of the argument? If we multiply 10+7 sqrt(2) by some constant C, the norm of the resulting expression will be 2 C^2. If we want the norm to be 2^3 = 8, then C^2 = 4 and C = 2 (or -2). When we then set [C(10+7 sqrt(2)) ]^(1/3) =[20+14 sqrt(2) ]^(1/3) = a + b sqrt(2), the corresponding norm will be a^2 - 2 b^2 = 2. The RHS of this expression, 2, is rational. But this doesn't mean that both a and b are also rational -- just that they might be rational. Alternative, we could have selected a new norm of 1 by requiring 2 C^2 = 1. Then C = 1/sqrt(2) = sqrt(2)/2. We would then be seeking (5 sqrt(2) + 7)^(1/3) = a sqrt(2) + b with 2 a^2 - b^2 = 1. Again, the RHS, 1, is rational which means that the coefficients a and b might also be rational. Choosing a = b = 1 here gives the desired result for the cube root. Now, let's consider the nth root of 5 sqrt(2) + 7; that is we want to de-nest (5 sqrt(2) + 7)^(1/n) , where n is a positive integer. We investigate and find the norm is 1 (as found above). Because any power of 1 is 1 (but, of course, there are complex roots of unity), we can expect that (5 sqrt(2) + 7)^(1/n) = a sqrt(2) + b is true, but should not expect a and b to be rational in general. Is this correct?

[@philipfoy7117]

Wow! Had no technic to this, I just intuitively thought to 3rd power 2+√2 and then worked from there. It work interestingly.

[@Blaqjaqshellaq]

The solution can also be presented as 2^(1/6)*[1+2^(1/2)].

[@perkin524]

Baffling!

[@scottleung9587]

Nice!

[@yurenchu]

³√(10 + 7√2) = = ³√( [ 5√2 + 7 ]*√2 ) = ³√( [ (2+3)√2 + (6+1) ]*√2 ) = ³√( [ 2√2 + 3√2 + 3*2 + 1 ]*√2 ) = ³√( [ 2√2 + 3*2 + 3√2 + 1 ]*√2 ) = ³√( [ (√2)³ + 3(1)(√2)² + 3(1²)√2 + 1³ ]*√2 ) = ³√( [ (√2 + 1)³ ]*√2 ) = ³√( [ (√2 + 1) * ⁶√2 ]³ ) = (√2 + 1) * ⁶√2 = ³√(2²) + ⁶√2 = ³√4 + ⁶√2

[@hazalouldi7130]

very nice

[@simonrosin7818]

how to compare sqrt(8)^7 with sqrt(7)^8

[@yakupbuyankara5903]

(2+(2^(1/2))/(2^(1/3))^3

[@buzzybola]

I got 11/4 doing it in my head in less than a minute lol

[@Waffle_6]

no you didn’t

[@rakenzarnsworld2]

​​@@Waffle_6The actual answer is 27/10

[@CriticSimon]

You should be spending more time on your solutions! 😆

[@buzzybola]

@@CriticSimon nah man fast approximations > slow accuracy

[@forcelifeforce]

@@buzzybola -- *No, fast on attempting your solution equals a wrong answer.* "Haste makes waste." You have a garbage post.

[@GourangaPL]

Now that's one thing i still don't understand, when can you use that b = ak to get one equation from a system?

[@SyberMath]

@@GourangaPL when the system is homogeneous meaning this substitution upon division will yield an equation in a single variable. You can tell by the sum of the powers in each term (it’s always 3)

[@GourangaPL]

@@SyberMath ah i see, so the sum of powers is the same, can we do similar thing with for example 3 equations 3 variables if every term has the same sum of powers? like b = ak and c = al ?