设原式= (7+5√2)^1/3 - (5√2-7)^1/3 = m > 0
m³ = (7+5√2) - (5√2-7) - 3 [(7+5√2)²(5√2-7)]^1/3 + 3 [(7+5√2)(5√2-7)²]^1/3
= 14 - 3 [ (7+5√2)^1/3 - (5√2-7)^1/3 ]
= 14 - 3m
m³+3m-14 = 0
(m-2)(m²+2m+7)=0 ,m²+2m+7>0
∴ 原式=m=2
变式1:设原式= √(3+√5) - √(3-√5) = m > 0
m² = (3+√5) + (3-√5) - 2√[(3+√5) (3-√5)]
= 6 - 4 = 2
∴ 原式=m=√2
变式2:设原式= (20+14√2)^1/3 - (20-14√2)^1/3 = m >0
m³ = (20+14√2) - (20-14√2) - 3 [(20+14√2)²(20-14√2)]^1/3 + 3 [(20+14√2)(20-14√2)²]^1/3
= 28√2 - 6 [ (20+14√2)^1/3 - (20-14√2)^1/3 ]
m³+6m-28√2 = 0
(m-2√2)[m²+(6+2√2)m+14]=0
∴ 原式=m=2√2
