Let's start with the quadratic equation:
[tex]\[ x^2 - 4k\sqrt{2} x + 2k^4 - 1 = 0 \][/tex]
where [tex]\( k \)[/tex] is a positive constant, and the roots are given as [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex].
### Step 1: Relation between the roots and coefficients
Using Vieta's formulas for the quadratic equations:
1. Sum of the roots ([tex]\(\alpha + \beta\)[/tex]):
[tex]\[ \alpha + \beta = 4k\sqrt{2} \][/tex]
2. Product of the roots ([tex]\(\alpha\beta\)[/tex]):
[tex]\[ \alpha \beta = 2k^4 - 1 \][/tex]
### Step 2: Given condition on the sum of squares of the roots
We are given:
[tex]\[ \alpha^2 + \beta^2 = 66 \][/tex]
Using the identity [tex]\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)[/tex], we substitute the values obtained from Vieta's formulas:
[tex]\[ \alpha^2 + \beta^2 = (4k\sqrt{2})^2 - 2(\alpha\beta) \][/tex]
[tex]\[ 66 = 16k^2 \cdot 2 - 2(2k^4 - 1) \][/tex]
[tex]\[ 66 = 32k^2 - 4k^4 + 2 \][/tex]
[tex]\[ 64 = 32k^2 - 4k^4 \][/tex]
[tex]\[ 4k^4 - 32k^2 + 64 = 0 \][/tex]
### Step 3: Solve for [tex]\(k\)[/tex]
Divide the entire equation by 4:
[tex]\[ k^4 - 8k^2 + 16 = 0 \][/tex]
This can be treated as a quadratic in terms of [tex]\( k^2 \)[/tex]. Let [tex]\( y = k^2 \)[/tex]:
[tex]\[ y^2 - 8y + 16 = 0 \][/tex]
Solve the quadratic equation:
[tex]\[ y = \frac{8 \pm \sqrt{64 - 64}}{2} = \frac{8 \pm 0}{2} = 4 \][/tex]
Thus:
[tex]\[ k^2 = 4 \][/tex]
[tex]\[ k = 2 \][/tex]
(Because [tex]\( k \)[/tex] is a positive constant)
### Step 4: Finding [tex]\( \alpha^3 + \beta^3 = p\sqrt{2} \)[/tex]
We use the identity:
[tex]\[ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha^2 + \beta^2) - \alpha\beta) \][/tex]
Substitute the known values:
[tex]\[ \alpha + \beta = 4k\sqrt{2} = 4 \cdot 2 \sqrt{2} = 8\sqrt{2} \][/tex]
[tex]\[ \alpha^2 + \beta^2 = 66 \][/tex]
[tex]\[ \alpha\beta = 2k^4 - 1 = 2 \cdot 2^4 - 1 = 31 \][/tex]
Then:
[tex]\[ \alpha^3 + \beta^3 = (8\sqrt{2}) \left(66 - 31\right) \][/tex]
[tex]\[ \alpha^3 + \beta^3 = 8\sqrt{2} \cdot 35 \][/tex]
[tex]\[ \alpha^3 + \beta^3 = 280\sqrt{2} \][/tex]
Since it is given that [tex]\(\alpha^3 + \beta^3 = p\sqrt{2}\)[/tex], we compare:
[tex]\[ 280\sqrt{2} = p\sqrt{2} \][/tex]
Thus:
[tex]\[ p = 280 \][/tex]
The value of [tex]\( p \)[/tex] is:
[tex]\[ \boxed{280} \][/tex]
