Question Number 103286 by bemath last updated on 14/Jul/20
Commented bybemath last updated on 14/Jul/20
Answered by floor(10²Eta[1]) last updated on 14/Jul/20
Answered by OlafThorendsen last updated on 14/Jul/20
![x = ((1/9))^(1/3) −((2/9))^(1/3) + ((4/9))^(1/3) x = (1/3^(2/3) )(1−2^(1/3) +2^(2/3) ) x^3 = (1/9)(1−2^(1/3) +2^(2/3) )^3 (a+b+c)^3 = a^3 +b^3 +c^3 +3(a+b)(b+c)(c+a) a = 1, b = −2^(1/3) , c = 2^(2/3) x^3 = (1/9)[1−2+4+3(1−2^(1/3) )(−2^(1/3) +2^(2/3) )(2^(2/3) +1)] x^3 = (1/9)[3+3.2^(1/3) (1−2^(1/3) )(2^(1/3) −1)(2^(2/3) +1)] x^3 = (1/9)[3−3.2^(1/3) (2^(2/3) −2.2^(1/3) +1)(2^(2/3) +1)] x^3 = (1/3)[1+2^(1/3) (2^(4/3) −2^(2/3) −1)(2^(2/3) +1)] x^3 = (1/3)[1+(2^(5/3) −2−2^(1/3) )(2^(2/3) +1)] x^3 = (1/3)[1+(2^(7/3) −2^(5/3) −2+2^(5/3) −2−2^(1/3) )] x^3 = (1/3)[−3+2^(7/3) −2^(1/3) ] x^3 = (1/3)[−3+4.2^(1/3) −2^(1/3) ] x^3 = (1/3)[−3+3.2^(1/3) ] x^3 = 2^(1/3) −1 x = (((2)^(1/3) −1))^(1/3)](Q103291.png)
Answered by bramlex last updated on 14/Jul/20
![consider : x^2 −xy+y^2 = ((x^3 +y^3 )/(x+y)) with x = ((2/3))^(1/3) , y=((1/3))^(1/3) ((4/9))^(1/3) −((2/9))^(1/3) +((1/3))^(1/3) = (1/(((2/3))^(1/3) +((1/3))^(1/3) )) = (1/(((1/3))^(1/3) ((2)^(1/3) +1))) = ((3)^(1/3) /((2)^(1/3) +1)) [ recall (x+y)^3 =x^3 +3x^2 y+3xy^2 +y^3 ] = ((3/(((2)^(1/3) +1)^3 )))^(1/3) = ((3/(3+3 (2)^(1/3) +3((2)^(1/3) )^2 )))^(1/3) = ((1/(1+(2)^(1/3) +((2)^(1/3) )^2 )))^(1/3) =((1/((((2−1)/((2)^(1/3) −1))))))^(1/3) = (( (2)^(1/3) −1))^(1/3)](Q103297.png)