1. **Problem statement:** Find the values of $a$ and $b$ for the cubic curve $$y = x^3 + ax^2 + bx + 3$$ given that it passes through the point $(1,8)$ and the tangent line at that point is $$y = 2x + 6.$$ 2. **Step 1: Use the point on the curve.** Substitute $x=1$ and $y=8$ into the cubic equation: $$8 = 1^3 + a(1)^2 + b(1) + 3 = 1 + a + b + 3 = a + b + 4.$$ Simplify to get: $$a + b = 4.$$ 3. **Step 2: Use the slope of the tangent line.** The derivative of the cubic curve is: $$y' = 3x^2 + 2ax + b.$$ At $x=1$, the slope of the tangent line is given by the derivative: $$y'(1) = 3(1)^2 + 2a(1) + b = 3 + 2a + b.$$ The slope of the tangent line $y = 2x + 6$ is 2, so: $$3 + 2a + b = 2.$$ Simplify: $$2a + b = 2 - 3 = -1.$$ 4. **Step 3: Solve the system of equations:** From step 2 and 3, we have: $$\begin{cases} a + b = 4 \\ 2a + b = -1 \end{cases}$$ Subtract the first equation from the second: $$\cancel{2a} + b - (\cancel{a} + b) = -1 - 4$$ $$a = -5.$$ Substitute $a = -5$ into $a + b = 4$: $$-5 + b = 4 \implies b = 9.$$ 5. **Final answer:** $$a = -5, \quad b = 9.$$