設a，b為實數，若a-b=6，ab=3，求a五方+b五方=?

方法一：

a - b =6
(a - b)² = 6²
a² - 2ab + b² = 36 …… [1]

ab = 3
4ab = 12 …… [2]

[1] + [2]：
a² + 2ab + b² = 48
(a + b)² = 48
a + b = ±√48
a + b = ±4√3

因為 (a + b)³ = a³ + 3a²b + 3ab² + b³
所以 a³ + b³
= (a + b)³ - 3a²b - 3ab²
= (a + b)³ - 3ab(a + b)

因為 (a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ - 5ab⁴ + b⁶
所以，a⁵ + b⁵
= (a + b)⁵ - 5a⁴b - 10a³b² - 10a²b³ - 5ab⁴
= (a + b)⁵ - 5a⁴b - 5ab⁴ - 10a³b² - 10a²b³
= (a + b)⁵ - 5ab(a³ + b³) - 10a²b²(a + b)
= (a + b)⁵ - 5ab[(a + b)³ - 3ab(a + b)] - 10a²b²(a + b)
= (a + b)⁵ - 5ab[(a + b)³ - 3ab(a + b)] - 10(ab)²(a + b)
= (±4√3)⁵ - 5×3×[(±4√3)³ - 3×3×(±4√3)] - 10×(3)²×(±4√3)
= ±9216√3 - (±2340√3) - (±360√3)
= ±6516√3

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方法二：

a - b = 6 …… [1]
ab = 3 …… [2]

由 [1]：
a = b + 6 …… [3]

將 [3] 代入 [2] 中：
(b + 6)b = 3
b² + 6b - 3 = 0
b = [-6 ± √(6² + 4*3)] / 2
b = -3 ± 2√3

將b = -3 ± 2√3 代入 [3] 中：
a = 3 ± 2√3

a⁵ = (3 ± 2√3)⁵
a⁵ = 3⁵ ± 5×3⁴×(2√3) ± 10×3³×(2√3)² ± 10×3²×(2√3)³ ± 5×3×(2√3)⁴ ± (2√3)⁵
a⁵ = 243 + (±810√3) + (±3240) + (±2160√3) + (±2160) + (±288√3)] …… [4]

b⁵ = (-3 ± 2√3)⁵
b ⁵ = (-3)⁵ ± 5×(-3)⁴×(2√3) ± 10×(-3)³×(2√3)² ± 10×(-3)²×(2√3)³ ± 5×(-3)×(2√3)⁴ ± (2√3)⁵
b⁵ = -243 + (±810√3) - (±3240) + (±2160√3) - (±2160) + (±288√3)] …… [5]

[4]+[5]：
a⁵ + b⁵ = 2×(±810√3) + 2×(±2160√3) + 2×(±288√3)]
a⁵ + b⁵ = ±6516√3

Sol
Set p=a+b
(a+b)+(a-b)=2a=p+6
(a+b)-(a-b)=2b=p-6
4ab=p^2-36=12
p^2=48
p=+/-4√3
b=a－6
ab=a(a－6)=3
a^2=6a+3
a^3=6a^2+3a=6(6a+3)+3a=39a+18
a^4=39a^2+18a=39(6a+3)+18a=252a+117
a^5=252a^2+117a=252(6a+3)+117a=1629a+756
a=b+6
ab=b(a+6)=3
b^2=－6b+3
b^3=－6b^2+3b=－6(－6b+3)+3b=39b－18
b^4=39b^2－18b=39(－6b+3)－18b=－252b+117
b^5=－252(－6b+3)+117b=1629b－756
a^5+b^5=1629(a+b)=1629p=+/-6516√3

(a+b)^2-4ab=36
(a+b)^2=48
(a+b)=+/-48^(1/2)
(a^3+b^3)=(a+b)(a^2-ab+b^2)
a^2+b^2=(a+b)^2-2ab
a^5+b^5=(a^3+b^3)(a^2+b^2)-a^2b^2(a+b)
=(a+b)[(a+b)^2-3ab)][(a+b)^2-2ab)]-(ab)^2(a+b)
=[+/-48^(1/2)](48-9)(48-6)-(9)[+/-48^(1/2)]
=1638[+/-48^(1/2)]-9[+/-48^(1/2)]
=1629[+/-4(3)^1/2]
=+/-6516(3)^1/2