To prove that \( P(y) = y^2 \) for all \( y \in \mathbb{R} \), we will analyze the polynomial \( P(x) \) given by:
\[
P(x) = \frac{(x-a)(x-b)}{(c-a)(c-b)}c^2 + \frac{(x-b)(x-c)}{(a-b)(a-c)}a^2 + \frac{(x-c)(x-a)}{(b-c)(b-a)}b^2
\]
### Step 1: Evaluate \( P(a) \)
Substituting \( x = a \):
\[
P(a) = \frac{(a-a)(a-b)}{(c-a)(c-b)}c^2 + \frac{(a-b)(a-c)}{(a-b)(a-c)}a^2 + \frac{(a-c)(a-a)}{(b-c)(b-a)}b^2
\]
The first and third terms vanish because they contain \( (a-a) \) and \( (a-a) \). Thus, we have:
\[
P(a) = a^2
\]
### Step 2: Evaluate \( P(b) \)
Substituting \( x = b \):
\[
P(b) = \frac{(b-a)(b-b)}{(c-a)(c-b)}c^2 + \frac{(b-b)(b-c)}{(a-b)(a-c)}a^2 + \frac{(b-c)(b-a)}{(b-c)(b-a)}b^2
\]
Again, the first and second terms vanish. Thus, we have:
\[
P(b) = b^2
\]
### Step 3: Evaluate \( P(c) \)
Substituting \( x = c \):
\[
P(c) = \frac{(c-a)(c-b)}{(c-a)(c-b)}c^2 + \frac{(c-b)(c-c)}{(a-b)(a-c)}a^2 + \frac{(c-c)(c-a)}{(b-c)(b-a)}b^2
\]
The second and third terms vanish. Thus, we have:
\[
P(c) = c^2
\]
### Step 4: Conclude the polynomial form of \( P(x) \)
Since \( P(a) = a^2 \), \( P(b) = b^2 \), and \( P(c) = c^2 \), we can conclude that \( P(x) \) is a quadratic polynomial that passes through the points \( (a, a^2) \), \( (b, b^2) \), and \( (c, c^2) \).
### Step 5: Form of \( P(x) \)
Since \( P(x) \) is a quadratic polynomial and we have three points, we can express \( P(x) \) in the form:
\[
P(x) = k(x^2) + mx + n
\]
### Step 6: Compare coefficients
Since \( P(x) \) must equal \( x^2 \) at three distinct points \( a \), \( b \), and \( c \), we can conclude that:
- The coefficient of \( x^2 \) must be 1.
- The coefficients of \( x \) and the constant term must be 0.
Thus, we have:
\[
P(x) = x^2
\]
### Conclusion
Therefore, we have shown that \( P(y) = y^2 \) for all \( y \in \mathbb{R} \).
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