If `a ,b , in R` such that `a+b=1a n d(1-2a b0(a 63+b^3)=12` . The value of `(a^2+b^2)` is equal to____.

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Set up the equations
We are given two equations:
1. \( a + b = 1 \)
2. \( (1 - 2ab)(a^3 + b^3) = 12 \)

### Step 2: Express \( a^2 + b^2 \) in terms of \( a + b \) and \( ab \)
We know that:
\[ a^2 + b^2 = (a + b)^2 - 2ab \]
Substituting \( a + b = 1 \):
\[ a^2 + b^2 = 1^2 - 2ab = 1 - 2ab \]
Let \( x = a^2 + b^2 \). Thus, we have:
\[ x = 1 - 2ab \quad \text{(1)} \]

### Step 3: Find \( a^3 + b^3 \) using the identity
Using the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
We can rewrite \( a^2 - ab + b^2 \) as:
\[ a^2 + b^2 - ab = x - ab \]
Thus:
\[ a^3 + b^3 = (a + b)(a^2 + b^2 - ab) = 1(x - ab) = x - ab \quad \text{(2)} \]

### Step 4: Substitute into the second equation
Now, substitute \( a^3 + b^3 \) from equation (2) into the second equation:
\[ (1 - 2ab)(x - ab) = 12 \]
Expanding this gives:
\[ x - ab - 2abx + 2(ab)^2 = 12 \]
Rearranging:
\[ x - ab - 2abx + 2(ab)^2 - 12 = 0 \quad \text{(3)} \]

### Step 5: Substitute \( ab \) from equation (1)
From equation (1), we have \( ab = \frac{1 - x}{2} \). Substitute this into equation (3):
\[ x - \frac{1 - x}{2} - 2\left(\frac{1 - x}{2}\right)x + 2\left(\frac{1 - x}{2}\right)^2 - 12 = 0 \]
Simplifying this will lead to a quadratic equation in \( x \).

### Step 6: Solve the quadratic equation
After simplifying, we get:
\[ x^2 - x - 24 = 0 \]
Factoring gives:
\[ (x - 3)(3x + 8) = 0 \]
Thus, the solutions are:
1. \( x - 3 = 0 \) → \( x = 3 \)
2. \( 3x + 8 = 0 \) → \( x = -\frac{8}{3} \) (not valid since \( a^2 + b^2 \) must be non-negative)

### Conclusion
The only valid solution is:
\[ a^2 + b^2 = x = 3 \]

### Final Answer
The value of \( a^2 + b^2 \) is **3**.