Solution: a-1≥0, a≥1 introduces parameters x, y, z, sz=(a+8)/3√[(a-1)/3]x=³√(a+z)y=³√(a-z)s=x+y∴Original formula =s=x+y=³√(a+z)+³√(a-z)∴x³+y³=2a…①xy=³√(a²-z²)…② From ①:x³+y³=(x+y)(x²+y²-xy)=(

Solution: a-1≥0, a≥1

Introduce parameters x, y, z, s

z=(a+8)/3√[(a-1)/3]

x=³√(a+z)

y=³√(a-z)

s=x+y

s=x+y

∴Original formula=s=x+y=³√(a+z)+³√(a-z)

∴x³+y³=2a…①

xy=³√(a²-z²)…②

from ①:x³+y³=(x+y)(x²+y²-xy)=(x+y)³-3xy(x+y)=(x+y)³-3xy(x+y)=(x+y)³-3³√(a²-z²)(x+y)=2a

∵a²-z²=a²-z²)(x+y)=2a

∵a²-z²=a²-z²)(x+y)=2a

∵a²-z²=a²-z²)(x+y)=2a

∵a²-z²=a²-(a²+16a+64)(a-1) /27

=a²-(a³-a²+16a²-16a²-16a+64a-64)/27

=-(a³-12a²+48a-64)/27

=-(a-4)³/27=(4-a)³/27

∴³√(a²-z²)=³√[(4-a)/3]³=(4-a)/3

∴(x+y)³+(a-4) (x+y)-2a=0

∴(x+y)³+(a-4) (x+y)-2a=0

html l0∵s=x+y

∴s³+(a-4)s-2a=0

∴(s³-8)+(a-4)s-2(a-4)=0

∴(s-2)(s²+2s+4)+(a-4)(s-2)=0

∴(s-2)(s²+2s+a)=0

∴s1=2, s2=-2+√(1-a), s3=-2-√(1-a)(a≥1)