✔ 最佳答案
∛5 - ∛4
= (5 - 4) / (∛5² + ∛5 ∛4 + ∛4²) ) ......... ∵ a - b = (a³ - b³) / (a² + ab + b²)
= 1 / (∛25 + ∛20 + 2∛2)Note that 25 * 20 * 2 = 1000 is a perfect cube.
To rationalize the denominator, assuming that
(∛25 + ∛20 + 2∛2) (a∛25 + b∛20 + c∛2)² is a rational number
while the numerator is a perfect square = (a∛25 + b∛20 + c∛2)² .
(∛25 + ∛20 + 2∛2) (a∛25 + b∛20 + c∛2)²= (∛25 + ∛20 + 2∛2) (a²∛625 + b²∛400 + c²∛4 + 2ab∛500 + 2bc∛40 + 2ca∛50)= (∛25 + ∛20 + 2∛2) (5a²∛5 + 2b²∛50 + c²∛4 + 10ab∛4 + 4bc∛5 + 2ca∛50)= (∛25 + ∛20 + 2∛2) ( (5a²+4bc)∛5 + (2b²+2ca)∛50 + (c²+10ab)∛4 )= (5a²+4bc)∛125 + (2b²+2ca)∛1250 + (c²+10ab)∛100
+ (5a²+4bc)∛100 + (2b²+2ca)∛1000 + (c²+10ab)∛80
+ 2(5a²+4bc)∛10 + 2(2b²+2ca)∛100 + 2(c²+10ab)∛8= 5(5a²+4bc) + 5(2b²+2ca)∛10 + (c²+10ab)∛100
+ (5a²+4bc)∛100 + 10(2b²+2ca) + 2(c²+10ab)∛10
+ 2(5a²+4bc)∛10 + 2(2b²+2ca)∛100 + 4(c²+10ab)= 5(5a²+4bc) + 10(2b²+2ca) + 4(c²+10ab)
+ ( 5(2b²+2ca) + 2(c²+10ab) + 2(5a²+4bc) )∛10
+ ( (c²+10ab) + (5a²+4bc) + 2(2b²+2ca) )∛100For it is rational ,
( 5(2b²+2ca) + 2(c²+10ab) + 2(5a²+4bc) )∛10 = 0
{
( (c²+10ab) + (5a²+4bc) + 2(2b²+2ca) )∛100 = 0 ⇒
5(b²+ca) + (c²+10ab) + (5a²+4bc) = 0 ... (1)
{
(c²+10ab) + (5a²+4bc) + 2(2b²+2ca) = 0 ... (2)(1) - (2) :
5(b²+ca) - 2(2b²+2ca) = 0
b² = - ac
Substitute into (1) or (2) :
(c²+10ab) + (5a²+4bc) = 0 ... (3)Let a = - bk , c = b/k , substitute into (3) :(b²/k² - 10b²k) + (5b²k² + 4b²/k) = 0
b² (1/k² - 10k + 5k² + 4/k) = 0
1/k² - 10k + 5k² + 4/k = 0 since b² ≠ 0 otherwise a = b = c = 0 ,
lead the denominator = 0.5k⁴- 10k³ + 4k + 1 = 0
(k - 1) (5k³ - 5k² - 5k - 1) = 0 k = 1 or k = 1.67035674... (irrational , ignore it)Put k = 1 , then a = - b , c = b ,
i.e. - a = b = cSet - a = b = c = 1 ,
the denominator
= 5(5a²+4bc) + 10(2b²+2ca) + 4(c²+10ab)
= 5(5 + 4) + 10(2 - 2) + 4(1 - 10)
= 9∴ 1 / (∛25 + ∛20 + 2∛2) = (- ∛25 + ∛20 + ∛2)² / 9
then
√(∛5 - ∛4)
= 1 / √(∛25 + ∛20 + 2∛2)
= (- ∛25 + ∛20 + ∛2) / 3