Let `alpha , beta` be non-real roots of `(x^(2) + x - 3) (x^(2) + x - 2) - 12 = 0 " then" alpha ^(73) + beta^(70)`= ________

Answer

To solve the problem, we need to find the value of \( \alpha^{73} + \beta^{70} \) where \( \alpha \) and \( \beta \) are the non-real roots of the equation:

\[ (x^2 + x - 3)(x^2 + x - 2) - 12 = 0 \]

### Step 1: Simplify the equation

Let \( y = x^2 + x \). Then we can rewrite the equation as:

\[ (y - 3)(y - 2) - 12 = 0 \]

Expanding this gives:

\[ y^2 - 5y + 6 - 12 = 0 \]

This simplifies to:

\[ y^2 - 5y - 6 = 0 \]

### Step 2: Solve for \( y \)

Now, we can use the quadratic formula to find the roots of the equation \( y^2 - 5y - 6 = 0 \):

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1, b = -5, c = -6 \):

\[ y = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ y = \frac{5 \pm \sqrt{25 + 24}}{2} = \frac{5 \pm \sqrt{49}}{2} = \frac{5 \pm 7}{2} \]

This gives us:

\[ y_1 = \frac{12}{2} = 6 \quad \text{and} \quad y_2 = \frac{-2}{2} = -1 \]

### Step 3: Find \( x \) values

Now we substitute back for \( y \):

1. For \( y = 6 \):
\[ x^2 + x - 6 = 0 \]
Factoring gives:
\[ (x - 2)(x + 3) = 0 \implies x = 2 \quad \text{or} \quad x = -3 \]

2. For \( y = -1 \):
\[ x^2 + x + 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{-1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]
Thus, the non-real roots are:
\[ \alpha = \frac{-1 + i\sqrt{3}}{2}, \quad \beta = \frac{-1 - i\sqrt{3}}{2} \]

### Step 4: Recognize \( \alpha \) and \( \beta \) as cube roots of unity

Notice that \( \alpha \) and \( \beta \) can be expressed in terms of the cube roots of unity:
\[ \alpha = \omega, \quad \beta = \omega^2 \]
where \( \omega = \frac{-1 + i\sqrt{3}}{2} \) and \( \omega^2 = \frac{-1 - i\sqrt{3}}{2} \).

### Step 5: Calculate \( \alpha^{73} + \beta^{70} \)

Using properties of cube roots of unity:
\[ \alpha^{73} = \omega^{73} = \omega^{73 \mod 3} = \omega^1 = \omega \]
\[ \beta^{70} = (\omega^2)^{70} = \omega^{140} = \omega^{140 \mod 3} = \omega^2 \]

Thus:
\[ \alpha^{73} + \beta^{70} = \omega + \omega^2 \]

### Step 6: Use the property of cube roots of unity

We know that:
\[ 1 + \omega + \omega^2 = 0 \implies \omega + \omega^2 = -1 \]

### Final Answer

Therefore, the value of \( \alpha^{73} + \beta^{70} \) is:

\[ \boxed{-1} \]