Solve the following problems: (a) Find x in the proportion p3 + pq + q2 : x :: p3 - q3 : (p-q)2. (b) Resolve into partial fraction: (3x+7)/((x+3)(x2+4)).

You're on a roll — Here are the solutions to the questions: 6. (a) Find x in the following proportion: p^3 + pq + q^2 : x :: p^3 - q^3 : (p-q)^2 Step 1: Write the proportion as an equation. (p^3 + pq + q^2)/(x) = (p^3 - q^3)/((p-q)^2) Step 2: Solve for x. x = ((p^3 + pq + q^2)(p-q)^2)/(p^3 - q^3) Step 3: Use the difference of cubes formula, p^3 - q^3 = (p-q)(p^2 + pq + q^2). Substitute this into the expression for x. x = ((p^2 + pq + q^2)(p-q)^2)/((p-q)(p^2 + pq + q^2)) Step 4: Cancel out common terms. x = ((p-q)^2)/(p-q) x = p-q The value of x is p-q. 6. (b) Resolve into partial fraction: (3x+7)/((x+3)(x^2+4)) Step 1: Set up the partial fraction decomposition. Since (x^2+4) is an irreducible quadratic factor, its numerator will be a linear expression. (3x+7)/((x+3)(x^2+4)) = (A)/(x+3) + (Bx+C)/(x^2+4) Step 2: Multiply both sides by (x+3)(x^2+4) to clear the denominators. 3x+7 = A(x^2+4) + (Bx+C)(x+3) Step 3: Solve for A, B, and C. To find A, substitute x=-3: 3(-3)+7 = A((-3)^2+4) + (B(-3)+C)(-3+3) -9+7 = A(9+4) + 0 -2 = 13A A = -(2)/(13) Expand the equation: 3x+7 = Ax^2+4A + Bx^2+3Bx+Cx+3C 3x+7 = (A+B)x^2 + (3B+C)x + (4A+3C) Compare coefficients: Coefficient of x^2: 0 = A+B Since A = -(2)/(13), then 0 = -(2)/(13) + B B = (2)/(13). Coefficient of x: 3 = 3B+C Substitute B = (2)/(13): 3 = 3((2)/(13)) + C 3 = (6)/(13) + C C = 3 - (6)/(13) = (39-6)/(13) = (33)/(13) (Alternatively, check constant term: 7 = 4A+3C. 7 = 4(-(2)/(13)) + 3((33)/(13)) 7 = -(8)/(13) + (99)/(13) 7 = (91)/(13) 7 = 7. This confirms the values of A, B, and C.) Step 4: Substitute the values of A, B, and C back into the decomposition. (3x+7)/((x+3)(x^2+4)) = (-2)/(13)x+3 + (2)/(13)x+(33)/(13)x^2+4 (3x+7)/((x+3)(x^2+4)) = -(2)/(13(x+3)) + (2x+33)/(13(x^2+4)) The partial fraction decomposition is -(2)/(13(x+3)) + (2x+33)/(13(x^2+4)). What's next?