Question Number 103286 by bemath last updated on 14/Jul/20

((1/9))^(1/3) −((2/9))^(1/3) + ((4/9))^(1/3)  = ?

Commented bybemath last updated on 14/Jul/20

yes. thank both

Answered by floor(10²Eta[1]) last updated on 14/Jul/20

(((2)^(1/3) −1))^(1/3)

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)

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)