Question

（x+45)的1/3次方减（x-16)的1/3次方等于1,求 （X1+ X2）

Answer 1

(X+45)-(X-16)=[(X+45)^1/3-(x-16)^1/3])[(X+45)^2/3+(X+45)^1/3(x-16)^1/3+(x-16)^2/3]

=[(X+45)^2/3+(X+45)^1/3(x-16)^1/3+(x-16)^2/3]

=[(X+45)^1/3-(x-16)^1/3]^2+3(X+45)^1/3(x-16)^1/3=61

(X+45)^1/3(x-16)^1/3=20

(X+45)(x-16)=8000,X1+ X2=-29