Solution
We are given the quadratic equation:
$$ x^2 - 4k\sqrt{2}\,x + 2k^4 - 1 = 0 $$
where $k$ is a positive constant and the roots are $\alpha$ and $\beta$.
Step 1: Use Vieta's Formulas
From Vieta’s formulas, for the quadratic equation $x^2 + bx + c = 0$, the sum and product of the roots are given by:
$$ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} $$
For the equation $x^2 - 4k\sqrt{2}\,x + (2k^4 - 1) = 0$, we have:
$$ \alpha + \beta = 4k\sqrt{2}, \quad \alpha\beta = 2k^4 - 1 $$
Step 2: Use the Identity for $ \alpha^2 + \beta^2 $
We are given that:
$$ \alpha^2 + \beta^2 = 66 $$
Recall the identity:
$$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta $$
Substitute the values from Vieta’s formulas:
$$ \alpha^2 + \beta^2 = (4k\sqrt{2})^2 - 2(2k^4 - 1) $$
Simplify:
$$ \alpha^2 + \beta^2 = 16k^2 \cdot 2 - 2(2k^4 - 1) $$
$$ 66 = 32k^2 - 2(2k^4 - 1) $$
$$ 66 = 32k^2 - 4k^4 + 2 $$
Now, simplify further:
$$ 66 - 2 = 32k^2 - 4k^4 $$
$$ 64 = 32k^2 - 4k^4 $$
Divide through by 4:
$$ 16 = 8k^2 - k^4 $$
Rearrange this to form a quadratic equation in $k^2$:
$$ k^4 - 8k^2 + 16 = 0 $$
Let $z = k^2$. The equation becomes:
$$ z^2 - 8z + 16 = 0 $$
Solve this quadratic equation using the quadratic formula:
$$ z = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(16)}}{2(1)} = \frac{8 \pm \sqrt{64 - 64}}{2} = \frac{8 \pm 0}{2} $$
Thus:
$$ z = 4 $$
Therefore:
$$ k^2 = 4, \quad k = 2 $$
Step 3: Find $p$
We are given that:
$$ \alpha^3 + \beta^3 = p\sqrt{2} $$
We use the identity for the sum of cubes:
$$ \alpha^3 + \beta^3 = (\alpha + \beta) \left( (\alpha + \beta)^2 - 3\alpha\beta \right) $$
Substitute the values for $\alpha + \beta$ and $\alpha\beta$:
$$ \alpha^3 + \beta^3 = (4k\sqrt{2}) \left( (4k\sqrt{2})^2 - 3(2k^4 - 1) \right) $$
Substitute $k = 2$:
$$ \alpha^3 + \beta^3 = (4(2)\sqrt{2}) \left( (4(2)\sqrt{2})^2 - 3(2(2)^4 - 1) \right) $$
$$ \alpha^3 + \beta^3 = 8\sqrt{2} \left( 64 \cdot 2 - 3(2(16) - 1) \right) $$
$$ \alpha^3 + \beta^3 = 8\sqrt{2} \left( 128 - 3(32 - 1) \right) $$
$$ \alpha^3 + \beta^3 = 8\sqrt{2} \left( 128 - 3 \cdot 31 \right) $$
$$ \alpha^3 + \beta^3 = 8\sqrt{2} \left( 128 - 93 \right) $$
$$ \alpha^3 + \beta^3 = 8\sqrt{2} \cdot 35 $$
$$ \alpha^3 + \beta^3 = 280\sqrt{2} $$
Thus, we have:
$$ p = 280 $$
Final Answer:
$$ p = 280 $$
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