If `alphaandbeta` are the roots of the quadratic equation `x^(2)-4x-6=0`, find the vlaues of
`alpha^(3)+beta^(3)`

Answer

To find the value of \( \alpha^3 + \beta^3 \) where \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 - 4x - 6 = 0 \), we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the form \( ax^2 + bx + c = 0 \).
Here, \( a = 1 \), \( b = -4 \), and \( c = -6 \).

### Step 2: Calculate the sum and product of the roots
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{-4}{1} = 4 \).
- The product of the roots \( \alpha \beta = \frac{c}{a} = \frac{-6}{1} = -6 \).

### Step 3: Use the identity for \( \alpha^3 + \beta^3 \)
We can use the identity:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2) \]
We need to find \( \alpha^2 + \beta^2 \) first.

### Step 4: Calculate \( \alpha^2 + \beta^2 \)
Using the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \]
Substituting the values we found:
\[ \alpha^2 + \beta^2 = (4)^2 - 2(-6) = 16 + 12 = 28 \]

### Step 5: Substitute into the identity for \( \alpha^3 + \beta^3 \)
Now we can substitute \( \alpha^2 + \beta^2 \) back into the identity:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha^2 + \beta^2) - \alpha \beta) \]
Substituting the known values:
\[ \alpha^3 + \beta^3 = 4(28 - (-6)) = 4(28 + 6) = 4(34) = 136 \]

### Final Answer
Thus, the value of \( \alpha^3 + \beta^3 \) is \( 136 \).

---