Okay, so I need to find the degree of the polynomial p(x) given by this expression:

p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)]

Hmm, the question says a, b, and c are distinct real numbers, so the denominators won't be zero. That's good. The polynomial is a sum of three terms, each of which is a rational function. But since p(x) is supposed to be a polynomial, each of these terms must combine to cancel out the denominators, right? So maybe each term simplifies to something with x squared in the numerator?

Wait, each term is a quadratic in the numerator divided by a constant. Let's check the degrees. Each term is of degree 2 in the numerator, divided by a constant, so each term is a quadratic polynomial. But when we add three quadratics together, the result could be a quadratic, a linear, or even a constant, depending on the coefficients. But since the problem states that a, b, c are distinct, maybe the coefficients are such that the higher degrees cancel out?

Alternatively, maybe there's a clever way to recognize this polynomial. It looks familiar. Let me think. It resembles Lagrange interpolation. Because in Lagrange interpolation, you have basis polynomials that are 1 at a particular node and 0 at others. Wait, in this case, the terms might be Lagrange basis polynomials for three points a, b, c.

Let me recall: The Lagrange interpolation formula for three points a, b, c would be L(x) = y_a * l_a(x) + y_b * l_b(x) + y_c * l_c(x), where each l_i(x) is the basis polynomial that's 1 at x_i and 0 at the others.

In our case, the coefficients in the numerator are (x - b)(x - c), (x - a)(x - c), and (x - a)(x - b). The denominators are (a - b)(a - c), (b - a)(b - c), and (c - a)(c - b). So if we think of this as the Lagrange basis polynomials evaluated at some points, then maybe it's interpolating a constant function?

Wait, suppose the function we are interpolating is a constant function, say f(x) = 1, then the Lagrange interpolating polynomial would be 1. So if p(x) is the Lagrange interpolation of f(x) = 1 at points a, b, c, then p(x) would be equal to 1 for all x. Let me check that.

If f(x) = 1, then the Lagrange interpolation formula would be:

p(x) = [ (x - b)(x - c) ] / [ (a - b)(a - c) ] + [ (x - a)(x - c) ] / [ (b - a)(b - c) ] + [ (x - a)(x - b) ] / [ (c - a)(c - b) ]

Which is exactly the expression given. So if we are interpolating the constant function 1 at three distinct points, then the interpolating polynomial must be the constant polynomial 1. Therefore, p(x) = 1 for all x. Hence, the degree of p(x) is 0.

Wait, but let me verify this. Let's test with specific values. Suppose a = 1, b = 2, c = 3. Then compute p(x):

First term: (x - 2)(x - 3) / [(1 - 2)(1 - 3)] = (x² - 5x + 6) / [(-1)(-2)] = (x² -5x +6)/2

Second term: (x - 1)(x - 3) / [(2 - 1)(2 - 3)] = (x² -4x +3) / [(1)(-1)] = - (x² -4x +3)

Third term: (x - 1)(x - 2) / [(3 - 1)(3 - 2)] = (x² -3x +2) / [(2)(1)] = (x² -3x +2)/2

Now sum them up:

First term: (1/2)x² - (5/2)x + 3

Second term: -x² +4x -3

Third term: (1/2)x² - (3/2)x +1

Adding them together:

(1/2 -1 +1/2)x² + (-5/2 +4 -3/2)x + (3 -3 +1)

Compute coefficients:

For x²: (1/2 -1 +1/2) = (1/2 +1/2) -1 = 1 -1 = 0

For x: (-5/2 +4 -3/2) = (-5/2 -3/2) +4 = (-8/2) +4 = -4 +4 = 0

For constants: 3 -3 +1 = 1

So p(x) = 0x² +0x +1 = 1. So in this case, p(x) is indeed 1, which is a constant. Therefore, the degree is 0.

But let me check another example to be sure. Let a=0, b=1, c=2.

First term: (x -1)(x -2)/[(0 -1)(0 -2)] = (x² -3x +2)/[(-1)(-2)] = (x² -3x +2)/2

Second term: (x -0)(x -2)/[(1 -0)(1 -2)] = x(x -2)/[1*(-1)] = -x(x -2) = -x² +2x

Third term: (x -0)(x -1)/[(2 -0)(2 -1)] = x(x -1)/[2*1] = (x² -x)/2

Sum them up:

First term: (1/2)x² - (3/2)x +1

Second term: -x² +2x

Third term: (1/2)x² - (1/2)x

Adding them together:

x² terms: 1/2 -1 +1/2 = (1/2 +1/2) -1 = 1 -1 = 0

x terms: -3/2 +2 -1/2 = (-3/2 -1/2) +2 = -2 +2 = 0

constants: 1

So again, p(x) = 1. Therefore, regardless of the values of a, b, c, as long as they are distinct, p(x) is the constant polynomial 1. Therefore, the degree is 0.

Alternatively, another approach: notice that each term is a quadratic, but when summed, the coefficients of x² and x cancel out, leaving a constant. Therefore, the degree is 0.

Wait, but let me think again. Suppose we have a polynomial of degree 2 added three times. Each term is degree 2, but when combined, the x² terms might cancel out. So if the coefficients of x² in each term sum to zero, then the x² term is eliminated, same for x. Then the constant terms must sum to a non-zero constant.

So let's compute the coefficient of x² in each term:

First term: (x - b)(x - c) = x² - (b + c)x + bc. Divided by (a - b)(a - c). So the coefficient of x² is 1 / [(a - b)(a - c)].

Second term: (x - a)(x - c) = x² - (a + c)x + ac. Divided by (b - a)(b - c). The coefficient of x² is 1 / [(b - a)(b - c)].

Third term: (x - a)(x - b) = x² - (a + b)x + ab. Divided by (c - a)(c - b). The coefficient of x² is 1 / [(c - a)(c - b)].

So total coefficient of x² in p(x):

1 / [(a - b)(a - c)] + 1 / [(b - a)(b - c)] + 1 / [(c - a)(c - b)]

Note that (b - a) = -(a - b), and (b - c) = -(c - b). So the second term becomes 1 / [-(a - b)(- (c - b))] = 1 / [(a - b)(c - b)]

Similarly, the third term: (c - a) = -(a - c), and (c - b) = -(b - c). So the third term becomes 1 / [-(a - c)(- (b - c))] = 1 / [(a - c)(b - c)]

Therefore, the coefficient of x² is:

1 / [(a - b)(a - c)] + 1 / [(a - b)(c - b)] + 1 / [(a - c)(b - c)]

Factor out 1/(a - b)(a - c)(b - c) from each term:

First term: 1 / [(a - b)(a - c)] = (b - c) / [(a - b)(a - c)(b - c)]

Second term: 1 / [(a - b)(c - b)] = (a - c) / [(a - b)(a - c)(b - c)]

Third term: 1 / [(a - c)(b - c)] = (a - b) / [(a - b)(a - c)(b - c)]

Therefore, adding them up:

[ (b - c) + (a - c) + (a - b) ] / [(a - b)(a - c)(b - c)]

Simplify numerator:

(b - c) + (a - c) + (a - b) = (b - c + a - c + a - b) = (2a - 2c) = 2(a - c)

Denominator: (a - b)(a - c)(b - c)

Therefore, the coefficient of x² is 2(a - c) / [(a - b)(a - c)(b - c)] = 2 / [(a - b)(b - c)]

Wait, but this is the coefficient of x². However, in our earlier examples, the coefficient of x² was zero. So there must be a mistake here. Wait, let me check the calculations again.

Wait, when I factored out 1/(a - b)(a - c)(b - c), the first term was:

1 / [(a - b)(a - c)] = 1 / [(a - b)(a - c)] = [1 * (b - c)] / [(a - b)(a - c)(b - c)]

Similarly, the second term:

1 / [(a - b)(c - b)] = 1 / [(a - b)(- (b - c))] = -1 / [(a - b)(b - c)] = [ - (a - c) ] / [(a - b)(a - c)(b - c)]

Wait, maybe my earlier step was incorrect.

Wait, let me re-express each term:

First term: 1 / [(a - b)(a - c)] = (b - c) / [(a - b)(a - c)(b - c)]

Second term: 1 / [(b - a)(b - c)] = 1 / [-(a - b)(b - c)] = -1 / [(a - b)(b - c)] = (a - c) / [(a - b)(a - c)(b - c)]

Third term: 1 / [(c - a)(c - b)] = 1 / [-(a - c)(- (b - c))] = 1 / [(a - c)(b - c)] = (a - b) / [(a - b)(a - c)(b - c)]

Therefore, adding them up:

[ (b - c) + (a - c) + (a - b) ] / [(a - b)(a - c)(b - c)]

Which is:

(b - c + a - c + a - b) / denominator

Simplify numerator:

b - c + a - c + a - b = (b - b) + (a + a) + (-c - c) = 0 + 2a - 2c = 2(a - c)

Therefore, the coefficient of x² is 2(a - c) / [(a - b)(a - c)(b - c)] = 2 / [(a - b)(b - c)]

Wait, but in the example earlier, when a=1, b=2, c=3, this coefficient would be 2 / [(1 - 2)(2 - 3)] = 2 / [(-1)(-1)] = 2 / 1 = 2. But in the example, the coefficient of x² was 0. So there must be a mistake here.

Wait, but in the example with a=1, b=2, c=3, the coefficient of x² in each term was 1/2, -1, and 1/2, so total 1/2 -1 +1/2 = -0, which is 0. But according to the formula above, it's 2 / [(1-2)(2-3)] = 2 / [(-1)(-1)] = 2. This is a contradiction. Therefore, my mistake must be in the calculation.

Wait, let's recompute the coefficient of x² in the first example.

First term: (x - 2)(x - 3)/( (1 - 2)(1 - 3) ) = (x² -5x +6)/ ( (-1)(-2) ) = (x² -5x +6)/2. So coefficient of x² is 1/2.

Second term: (x -1)(x -3)/( (2 -1)(2 -3) ) = (x² -4x +3)/ (1*(-1)) = -x² +4x -3. Coefficient of x² is -1.

Third term: (x -1)(x -2)/( (3 -1)(3 -2) ) = (x² -3x +2)/(2*1) = (x² -3x +2)/2. Coefficient of x² is 1/2.

Total coefficient of x²: 1/2 -1 +1/2 = 0. Which matches the example. But according to the formula I derived earlier, it's 2 / [ (a - b)(b - c) ] = 2 / [ (1 - 2)(2 - 3) ] = 2 / [ (-1)(-1) ] = 2. Which is 2, but in reality it's 0. So my general formula is wrong.

Therefore, my mistake is in the algebra when factoring out the terms. Let me try again.

Let me denote:

Term1: 1 / [(a - b)(a - c)]

Term2: 1 / [(b - a)(b - c)] = -1 / [(a - b)(b - c)]

Term3: 1 / [(c - a)(c - b)] = 1 / [ (a - c)(b - c) ]

So then, the total coefficient is:

Term1 + Term2 + Term3 = [1 / (a - b)(a - c)] - [1 / (a - b)(b - c)] + [1 / (a - c)(b - c)]

Factor terms:

First, note that (a - c) = -(c - a), (b - c) = -(c - b). So let's express all terms over a common denominator, which would be (a - b)(a - c)(b - c).

So:

Term1: 1 / [(a - b)(a - c)] = (b - c) / [(a - b)(a - c)(b - c)]

Term2: -1 / [(a - b)(b - c)] = - (a - c) / [(a - b)(a - c)(b - c)]

Term3: 1 / [(a - c)(b - c)] = (a - b) / [(a - b)(a - c)(b - c)]

Therefore, adding them:

[ (b - c) - (a - c) + (a - b) ] / [(a - b)(a - c)(b - c)]

Simplify numerator:

b - c - a + c + a - b = (b - c + c - a + a - b) = 0

Wait, that's zero! So the coefficient of x² is zero. Therefore, the coefficient of x² is zero. That matches the examples. So my initial mistake was in the algebra when expanding the terms. So in fact, the coefficient of x² is zero. Therefore, the x² terms cancel out.

Similarly, let's compute the coefficient of x.

First term: (x - b)(x - c) = x² - (b + c)x + bc. Coefficient of x is -(b + c). Divided by (a - b)(a - c). So coefficient is -(b + c)/[(a - b)(a - c)].

Second term: (x - a)(x - c) = x² - (a + c)x + ac. Coefficient of x is -(a + c). Divided by (b - a)(b - c). So coefficient is -(a + c)/[(b - a)(b - c)].

Third term: (x - a)(x - b) = x² - (a + b)x + ab. Coefficient of x is -(a + b). Divided by (c - a)(c - b). So coefficient is -(a + b)/[(c - a)(c - b)].

Total coefficient of x:

- [ (b + c)/[(a - b)(a - c)] + (a + c)/[(b - a)(b - c)] + (a + b)/[(c - a)(c - b)] ]

Again, let's express each term over a common denominator (a - b)(a - c)(b - c):

First term: -(b + c) / [(a - b)(a - c)] = - (b + c)(b - c) / [(a - b)(a - c)(b - c)]

Second term: -(a + c) / [(b - a)(b - c)] = - (a + c)(-1)(a - b) / [(a - b)(a - c)(b - c)] = (a + c)(a - b) / [(a - b)(a - c)(b - c)]

Third term: -(a + b) / [(c - a)(c - b)] = - (a + b)(-1)(a - c)(-1)(b - c) / [(a - b)(a - c)(b - c)] Wait, perhaps better to factor each term.

Wait, second term denominator: (b - a)(b - c) = -(a - b)(b - c). So:

Second term: -(a + c)/[ -(a - b)(b - c) ] = (a + c)/[ (a - b)(b - c) ] = (a + c)(a - c) / [ (a - b)(a - c)(b - c) ]

Third term denominator: (c - a)(c - b) = (a - c)(b - c). So:

Third term: -(a + b)/[ (a - c)(b - c) ] = - (a + b)(a - b) / [ (a - b)(a - c)(b - c) ]

Therefore, total coefficient of x:

[ - (b + c)(b - c) + (a + c)(a - c) - (a + b)(a - b) ] / [ (a - b)(a - c)(b - c) ]

Compute numerator:

- (b + c)(b - c) + (a + c)(a - c) - (a + b)(a - b)

First term: - (b² - c²)

Second term: (a² - c²)

Third term: - (a² - b²)

So total:

- b² + c² + a² - c² - a² + b² = 0

Therefore, the coefficient of x is 0. So both x² and x terms cancel out. Then the constant term.

Compute the constant terms:

First term: bc / [(a - b)(a - c)]

Second term: ac / [(b - a)(b - c)] = ac / [ - (a - b)(b - c) ] = -ac / [ (a - b)(b - c) ]

Third term: ab / [ (c - a)(c - b) ] = ab / [ (a - c)(b - c) ] = ab / [ (a - c)(b - c) ]

So total constant term:

bc / [(a - b)(a - c)] - ac / [ (a - b)(b - c) ] + ab / [ (a - c)(b - c) ]

Again, express over a common denominator (a - b)(a - c)(b - c):

First term: bc(b - c) / [(a - b)(a - c)(b - c)]

Second term: -ac(a - c) / [(a - b)(a - c)(b - c)]

Third term: ab(a - b) / [(a - b)(a - c)(b - c)]

So total numerator:

bc(b - c) - ac(a - c) + ab(a - b)

Let's expand each term:

First term: bc(b - c) = b²c - bc²

Second term: -ac(a - c) = -a²c + ac²

Third term: ab(a - b) = a²b - ab²

Combine all terms:

b²c - bc² - a²c + ac² + a²b - ab²

Let's group similar terms:

- a²c + a²b = a²(b - c)

+ b²c - ab² = b²(c - a)

+ ac² - bc² = c²(a - b)

So total:

a²(b - c) + b²(c - a) + c²(a - b)

This expression is known to be equal to (a - b)(b - c)(c - a). Let me verify:

(a - b)(b - c)(c - a) = (a - b)(b - c)(- (a - c)) = - (a - b)(b - c)(a - c)

But the numerator we have is a²(b - c) + b²(c - a) + c²(a - b). Let's check if this is equal to (a - b)(b - c)(c - a):

Expand (a - b)(b - c):

(a - b)(b - c) = ab - ac - b² + bc

Multiply by (c - a):

(ab - ac - b² + bc)(c - a)

= ab(c - a) - ac(c - a) - b²(c - a) + bc(c - a)

= abc - a²b - a c² + a²c - b²c + a b² + b c² - a b c

Simplify:

abc - a²b - a c² + a²c - b²c + a b² + b c² - a b c

= (-a²b + a²c) + (-a c² + b c²) + (-b²c + a b²) + (abc - abc)

= a²(c - b) + c²(b - a) + b²(a - c)

Which is same as:

a²(c - b) + b²(a - c) + c²(b - a)

But our numerator is a²(b - c) + b²(c - a) + c²(a - b) = - [a²(c - b) + b²(a - c) + c²(b - a) ]

Therefore, numerator = - (a - b)(b - c)(c - a) = (a - b)(b - c)(a - c)

Therefore, constant term:

[ (a - b)(b - c)(a - c) ] / [ (a - b)(a - c)(b - c) ] = 1

Therefore, the constant term is 1. Therefore, p(x) = 0x² + 0x +1 =1. Therefore, the degree is 0.

Therefore, the answer is \boxed{0}.

Alternatively, since the polynomial is the Lagrange interpolation polynomial for the constant function 1 at three distinct points, the interpolating polynomial must be the constant polynomial 1. Therefore, the degree is 0.

**Final Answer**
\boxed{0}
Therefore, the degree of the polynomial is $\boxed{0}$.
</think>