We are asked to find the sum of constants $a + b$ for a line $y = ax + b$ that is tangent to two curves $y = x^3$ and $y = 3x^2 - 4$ at the same point $x = t$.

Step 1: Condition for Tangency

For the line $y = ax + b$ to be tangent to both curves at $x = t$, it must satisfy two conditions:

1. The point of tangency: The line should pass through the points on both curves at $x = t$, meaning the $y$-values should match.

   • $t^3 = at + b$ (from $y = x^3$)
   • $3t^2 - 4 = at + b$ (from $y = 3x^2 - 4$)

2. Matching the slopes: The derivative of the curves at $x = t$ should equal the slope of the line, which is $a$.

   • Slope of $y = x^3$ at $x = t$: $3t^2$
   • Slope of $y = 3x^2 - 4$ at $x = t$: $6t$

Thus, the following system of equations emerges: $3t^2 = a$ $6t = a$ From this, we see $a = 3t^2 = 6t$.

Step 2: Solve for $t$

Equating the expressions for $a$: $3t^2 = 6t$ Dividing through by $t$ (assuming $t \neq 0$): $3t = 6 \quad \Rightarrow \quad t = 2$

Step 3: Find $a$ and $b$

Now that we know $t = 2$, we can substitute it back into the expressions for $a$ and solve for $b$: $a = 3t^2 = 3(2)^2 = 12$

Next, substitute $t = 2$ into either of the original point equations (say, $t^3 = at + b$) to find $b$: $2^3 = 12(2) + b$ $8 = 24 + b \quad \Rightarrow \quad b = -16$

Step 4: Calculate $a + b$

$a + b = 12 + (-16) = -4$

Thus, the value of $a + b$ is $\boxed{-4}$.

Would you like more details on any step, or further clarifications?

Here are 5 related questions for further practice:

1. What happens if the two curves have different slopes at the same point?
2. How do you find the point of tangency for a quadratic curve?
3. What is the general condition for a line to be tangent to a curve at a given point?
4. Can you find the value of $t$ if the curves are $y = x^3$ and $y = 5x^2 - 2$?
5. How would the answer change if the curve was $y = x^4$ instead of $y = x^3$?

Tip: Always verify the slopes when solving tangency problems involving multiple curves to ensure the tangency condition is met.