If α and β are roots of the equation, x² - 4√2kx + 2e4lnk = 0 for some k, and α² + β² = 66, then α³ + β³ is equal to

α + β = 4√2k, αβ = 2e4lnk
α² + β² = (α + β)² - 2αβ
66 = 16 × 2k² - 2(2e4lnk)
66 = 32k² - 4k⁴
4k⁴ - 32k² + 66 = 0
2k⁴ - 16k² + 33 = 0
k⁴ - 8k² + 33/2 = 0
Solving the quadratic equation for k², we get k² = 4 ± √(16 - 66/2) which is not real.
However, if the equation is x² - 4√2kx + 2e4lnk = 0, then αβ = 2k⁴
66 = 32k² - 4k⁴
4k⁴ - 32k² + 66 = 0
2k⁴ - 16k² + 33 = 0
(k² - 3)(2k² - 11) = 0
k² = 3 or k² = 11/2
Since k > 0, k = √3 or k = √(11/2)
Let's assume the equation is x² - 4√2kx + 2k⁴ = 0
Then α + β = 4√2k and αβ = 2k⁴
α² + β² = (α + β)² - 2αβ = 32k² - 4k⁴ = 66
4k⁴ - 32k² + 66 = 0
2k⁴ - 16k² + 33 = 0
(k² - 3)(2k² - 11) = 0
k² = 3 or k² = 11/2
If k² = 3, k = √3
If k² = 11/2, k = √(11/2)
(α + β)³ = α³ + β³ + 3αβ(α + β)
α³ + β³ = (α + β)³ - 3αβ(α + β)
If k = √3, α + β = 4√6, αβ = 18
α³ + β³ = (4√6)³ - 3(18)(4√6) = 384√6 - 216√6 = 168√6
If k = √(11/2), α + β = 4√11, αβ = 121/2
α³ + β³ = (4√11)³ - 3(121/2)(4√11) = 176√11 - 726√11 = -550√11
Let's assume the question has a typo and the equation is x² - 4√2x + 2k⁴ = 0
Then α + β = 4√2 and αβ = 2k⁴
(4√2)² - 2(2k⁴) = 66
32 - 4k⁴ = 66 which is not possible.
Let's assume the equation is x² - 4√2kx + 32 = 0
Then α + β = 4√2k and αβ = 32
(4√2k)² - 2(32) = 66
32k² - 64 = 66
32k² = 130
k² = 130/32 = 65/16
k = √(65/16)
(α + β)³ - 3αβ(α + β) = α³ + β³
(4√2k)³ - 3(32)(4√2k) = α³ + β³
128√2k³ - 384√2k = α³ + β³
If k = √(65/16), α³ + β³ = 128√2(65/16)√(65/16) - 384√2√(65/16) ≈ 280√2