(In reply to
re: Simply horrendous by Jer)
Interesting story. I've been very jet-lagged the last few days and find Perplexus is one way to cope with that.
I'd not attempted or even heard of 'rationalizing the denominator' before I read this problem. I looked at a few videos until I understood the basic idea enough to make some generalisations. Difference of squares is the starting point. I also remembered KS's trick from a while ago:
(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = a^3+b^3+c^3-3abc
(x-y)(x^2+xy+y^2) = (x^3-y^3), in this case (a^3+b^3+c^3)^3-(3abc)^3
and of course for higher powers there is always
1/(a^(1/(2x-1))+b^(1/(2x-1))) = 1*(R)/(a+b) where (a+b) = (a^x+b^x)/R,
e.g. (a+b) = (a^(3)+b^(3))/R, when R = a^2-ab+b^2.
This seems to be around the limit of the high-school toolkit for this type of puzzle. I'm guessing that a more mature approach would involve manipulating nth roots of unity, something well beyond my comfort zone.
After some effort I reached a complicated expression which clearly wasn't even close to a final answer. - the denominator was still
(99 + 70*2^(1/2) + 45*5^(1/5) + 30*2^(1/2)5^(1/5) + 51*5^(3/5) - 126*2^(1/2) 5^(3/5) + 5*5^(4/5)).
So I had a closer look at WolframAlpha which gives the expression I quoted as an 'alternate form', a part of the program I'd never really paid much attention to in the past. As a reality check I worked out the much simpler 1/(3^(1/3)+5^(1/5)):
Rationalise:
For the 5th power:
(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)=a^5+b^5
R(1)=(a^4-a^3b+a^2b^2-ab^3+b^4), a=(3^(1/3)), b=(5^(1/5))
Substituting 3*3^(1/3)-3*5^(1/5)+3^(2/3)5^(2/5)-3^(1/3)5^(3/5)+5^(4/5)
Now (3*3^(1/3) - 3*5^(1/5) + 3^(2/3)*5^(2/5) - 3^(1/3)*5^(3/5)+5^(4/5))/(3*3^(2/3)+5)
For the 3rd power:
a^3+b^3 = (a+b)(a^2-ab+b^2)
R(2) = a^2-ab+b^2 = (25 + 27*3^(1/3) - 15*3^(2/3))
For (3*3^(2/3)+5), ((3*3^(2/3))^3+5^3)= 368
Giving ((3*3^(1/3) - 3*5^(1/5) + 3^(2/3)*5^(2/5) - 3^(1/3)*5^(3/5)+5^(4/5)) (25 + 27*3^(1/3) - 15*3^(2/3)))/368
= 1/368*(25 + 27*3^(1/3) - 15*3^(2/3)) (3*3^(1/3) - 3*5^(1/5) + 3^(2/3)5^(2/5) - 3^(1/3) 5^(3/5) + 5^(4/5))
And indeed WolframAlpha gives the same result (in a different order) as the 'alternate form'. So I can't claim to have solved the problem personally but I did at least learn something.
Edited on January 20, 2019, 10:36 pm
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Posted by broll
on 2019-01-20 22:14:53 |
re(2): Simply horrendous
