answer:To find the coefficient of x^3 in the binomial expansion of (x+1)^n, we look at the general term in the expansion which is given by: T_{r+1} = {C_n}^r x^r (1)^{n-r}. To get the term containing x^3, we need to set r = 3. Plugging this into the general term, we get: T_{4} = {C_n}^3 x^3. Therefore, the coefficient of x^3 in the expansion is {C_n}^3, which represents the binomial coefficient and is defined as: {C_n}^3 = frac{n!}{3!(n-3)!}. Hence, the coefficient we're seeking is boxed{{C_n}^3}.

answer:First, we identify the greatest common factor (GCF) of the terms 56b^2 and 168b. The coefficients 56 and 168 have a GCF of 56, and the smallest power of b in both terms is b. Therefore, the GCF of 56b^2 and 168b is 56b. Now, factor 56b out of both terms: [ 56b^2 + 168b = 56b cdot b + 56b cdot 3 = 56b(b + 3) ] Thus, the factored form of the expression is boxed{56b(b+3)}.

answer:1. **Initial Setup**: Given a finite set A composed of positive integers, we need to find another finite set B such that A subseteq B and prod_{x in B} x = sum_{x in B} x^2. 2. **Defining the range for A**: Suppose A subseteq {1, 2, cdots, m} for some integer m geq 5. The condition m geq 5 ensures that k > 0 (as shown later). 3. **Constructing the set B**: To construct B, define the sequence { x_n } as follows: begin{align*} x_0 &= 1, x_k &= (m!) x_0 x_1 cdots x_{k-1} - 1 quad text{for} quad k geq 1. end{align*} 4. **Defining intermediate sets**: Let begin{align*} B_0 &= {1, 2, cdots, m}, B_i &= {1, 2, cdots, m, x_1, x_2, cdots, x_i}. end{align*} 5. **Setting up the equation for verification**: Define: k = m! - sum_{j=1}^{m} j^2. Since m geq 5, we know that k > 0, guaranteeing positive results in our calculation. 6. **Mathematical Induction**: We will prove by induction that for i geq 0, the following holds: prod_{x in B_i} x - sum_{x in B_i} x^2 = k - i. 7. **Base Case**: For i = 0: begin{align*} prod_{x in B_0} x &= 1 cdot 2 cdot cdots cdot m = m!, sum_{x in B_0} x^2 &= 1^2 + 2^2 + cdots + m^2. end{align*} Thus, prod_{x in B_0} x - sum_{x in B_0} x^2 = m! - sum_{j=1}^{m} j^2 = k. Hence, the base case holds true. 8. **Inductive Step**: Assume the statement holds for some i geq 0. We need to prove it for i+1: begin{align*} &prod_{x in B_{i+1}} x - sum_{x in B_{i+1}} x^2 &= ((m!) x_0 x_1 cdots x_i - 1) prod_{x in B_i} x - sum_{x in B_i} x^2 - ((m!) x_0 x_1 cdots x_i - 1)^2 &= ((m!) x_0 x_1 cdots x_i - 1) left( (m!) x_0 x_1 cdots x_i right) - sum_{x in B_i} x^2 - ((m!) x_0 x_1 cdots x_i - 1)^2 &= (m! x_0 x_1 cdots x_i) (m! x_0 x_1 cdots x_i - 1) - ((m!) x_0 x_1 cdots x_i - 1)^2 - sum_{x in B_i} x^2 &= (m! x_0 x_1 cdots x_i - 1) - sum_{x in B_i} x^2 &= k - (i + 1). end{align*} 9. **Conclusion**: By induction, it follows that for all i geq 0, prod_{x in B_i} x - sum_{x in B_i} x^2 = k - i. Thus, for i = k, we have: prod_{x in B_k} x - sum_{x in B_k} x^2 = 0, or prod_{x in B_k} x = sum_{x in B_k} x^2. Therefore, the original statement is verified and the solution is complete. blacksquare

answer:To find the coordinates of a point P(-2,3) symmetric to the origin in the Cartesian coordinate system, we use the property that the symmetric point of (x,y) with respect to the origin is (-x,-y). Applying this to point P: 1. Original coordinates of P: (-2, 3) 2. Applying symmetry with respect to the origin: (-(-2), -(3)) 3. Simplifying the coordinates: (2, -3) Therefore, the coordinates of the point symmetric to P(-2,3) with respect to the origin are (2, -3). Hence, the correct answer is boxed{C}.