A Nice Algebra Challenge | Radical Math

[@RyanLewis-Johnson-wq6xs]

Input
(16^(1/5) - 8^(1/5) + 4^(1/5) - 2^(1/5) + 1)/(256^(1/5) + 64^(1/5) + 16^(1/5) + 4^(1/5) + 1) = 2^(1/5) - 1
Result
True
Left hand side
(16^(1/5) - 8^(1/5) + 4^(1/5) - 2^(1/5) + 1)/(256^(1/5) + 64^(1/5) + 16^(1/5) + 4^(1/5) + 1)≈0.148698
Right hand side
2^(1/5) - 1≈0.148698
Logarithmic form
log(1 + 2 2^(1/5) + 2^(2/5) + 2 2^(3/5) + 2^(4/5), 16^(1/5) - 8^(1/5) + 4^(1/5) - 2^(1/5) + 1) - log(1 + 2 2^(1/5) + 2^(2/5) + 2 2^(3/5) + 2^(4/5), 256^(1/5) + 64^(1/5) + 16^(1/5) + 4^(1/5) + 1) = log(1 + 2 2^(1/5) + 2^(2/5) + 2 2^(3/5) + 2^(4/5), 2^(1/5) - 1)

[@RashmiRay-c1y]

Ley 2^1/5 = y and 4^1/5 = z = y^2. Then, the numerator in x, N = 1-y+y^2-y^3+y^4 = (y^5+1)/(y+1) and the denominator is D= 1+z+z^2+z^3+z^4 = (z^5-1)/(z-1) = [(y^5)^2-1]/(y^2-1). So x = N/D = (y-1)/(y^5-1) = 2^1/5-1. So, x+1 = 2^1/5. Again, E (to be evaluated) = 1/(x+1)^2 - 1/(x+1)^5 = 1/2^(2/5) - 1/2 > E = 1/2[8^(1/5) - 1].

[@RyanLewis-Johnson-wq6xs]

Input
((2^(1/5) - 1) ((2^(1/5) - 1) (2^(1/5) - 1) + 3 (2^(1/5) - 1) + 3))/(2^(1/5) - 1 + 1)^5 = 0.5(8^(1/5))- 0.5=x
Result
True
Left hand side
((2^(1/5) - 1) ((2^(1/5) - 1) (2^(1/5) - 1) + 3 (2^(1/5) - 1) + 3))/(2^(1/5) - 1 + 1)^5 = 1/2 (2^(3/5) - 1)
Right hand side
0.5 8^(1/5) - 0.5 = 1/2 (2^(3/5) - 1)
Logarithmic form
log_2(2^(1/5) - 1) + log_2((2^(1/5) - 1) (2^(1/5) - 1) + 3 (2^(1/5) - 1) + 3) - log_2((2^(1/5) - 1 + 1)^5) = log_2(0.5 8^(1/5) - 0.5)

[@mulla_modi]

I think now it would be useful to include problems from other areas of mathematics like geometry, permutation, probability, arithmetic, etc

[@abcekkdo3749]

E=(⁵√8-1)/2

[@Fjfurufjdfjd]

Χρησιμοποιω τις ταυτοτητες: α^5-1=(α-1)(α^4+α ^3+α^2+α+1)οπου α=2^(1/5) και β^5+1=(β+1)(β^4-β^3+β^2-β+1).εχω χ= 2^(1/5)-1. Τελικα Ε=Α/Π=[2^(3/5)-1]/2=[8^(1/5)-1]/2

[@Fjfurufjdfjd]

Οπου β=4^(1/5)

[@Fjfurufjdfjd]

Οπου β=4^(1/5)

[@Fjfurufjdfjd]

Ακυρο

[@RealQinnMalloryu4]

(x ➖ 2x+1). (x+1x+1),

[@RealQinnMalloryu4]

5^14^4 ➖ 5^12^3+5^12^2 ➖ 5^1^2^1+1^1/5^1^16^16+5^18^8+5^14^4+5^1^2^2 1^1 1^12^2^2^2+ ➖ 1^11^1+1^11^1 ➖ 1^1^1/1^14^4+1^12^32^3+1^12^22^2+1^11^1 1^1^1^1/2^2^2 1^1^1^1 1^1^1^1 1^1/1^1^1^2 1^2 (x ➖ 2x+1), {x^3+3x^2+3x}/5x^5 6x^6/5x^5 1.1x1^1 1^1.1^1x^1^1.1^1 1^1x^1^1 1x^1 (x ➖ 1x+1) .

[@RealQinnMalloryu4]

5^14^4 ➖ 5^12^3+5^1^2^2 ➖ 5^12^1+1^1/5^116^16+5^18^8+5^14^4+5^12^2+1^1 1^12^2^2+1^11^1+1^1^1^1 ➖ 1^11^1/1^14^4+1^12^32^3+1^12^22^2+1^11^1 1^11^1/2^2^2^2 1^1^1^1 1^1^1 1^1/1^1^1^2 1^2 (x ➖ 2x+1) {x^3+3x^2+3x}/5x^5 6x^6/5x^5 =1.1x1.
1 1^1.1^1x1^1.1^1 1^1x1^1 1x^1 (x ➖ 1x+1)

[@Quest3669]

? ={( 8^1/5)-1}/2