We are tasked with finding the value of $a^3b + ab^3$ given that $a - b = 5$ and $ab = 7$.

Step 1: Express $a$ and $b$ in terms of their sum or difference

Let: $S = a + b, \quad P = ab$ We know that: $a - b = 5 \quad \text{and} \quad ab = 7$ From the first equation: $a = b + 5$

Step 2: Substitute $a = b + 5$ into $ab = 7$

$(b + 5)b = 7$ $b^2 + 5b - 7 = 0$

Step 3: Solve the quadratic equation

The quadratic equation is: $b^2 + 5b - 7 = 0$ We use the quadratic formula: $b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ Here, $A = 1$, $B = 5$, and $C = -7$: $b = \frac{-5 \pm \sqrt{5^2 - 4(1)(-7)}}{2(1)}$ $b = \frac{-5 \pm \sqrt{25 + 28}}{2}$ $b = \frac{-5 \pm \sqrt{53}}{2}$

Thus: $b = \frac{-5 + \sqrt{53}}{2} \quad \text{or} \quad b = \frac{-5 - \sqrt{53}}{2}$

Step 4: Find $a$

Using $a = b + 5$:

1. For $b = \frac{-5 + \sqrt{53}}{2}$: $a = \frac{-5 + \sqrt{53}}{2} + 5 = \frac{-5 + \sqrt{53} + 10}{2} = \frac{5 + \sqrt{53}}{2}$

2. For $b = \frac{-5 - \sqrt{53}}{2}$: $a = \frac{-5 - \sqrt{53}}{2} + 5 = \frac{-5 - \sqrt{53} + 10}{2} = \frac{5 - \sqrt{53}}{2}$

Thus, the two pairs of $(a, b)$ are: $\left( \frac{5 + \sqrt{53}}{2}, \frac{-5 + \sqrt{53}}{2} \right) \quad \text{and} \quad \left( \frac{5 - \sqrt{53}}{2}, \frac{-5 - \sqrt{53}}{2} \right)$

Step 5: Compute $a^3b + ab^3$

Factor $a^3b + ab^3$ as: $a^3b + ab^3 = ab(a^2 + b^2)$

Compute $a^2 + b^2$

Using the identity: $a^2 + b^2 = (a + b)^2 - 2ab$ $a + b = S = 5 \quad \text{(from the sum of roots of the quadratic)} \quad \text{and} \quad ab = 7$ $a^2 + b^2 = 5^2 - 2(7) = 25 - 14 = 11$

Compute $a^3b + ab^3$

$a^3b + ab^3 = ab(a^2 + b^2) = 7(11) = 77$

Final Answer:

$a^3b + ab^3 = 77$

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Would you like further explanation or details?

Related Questions:

1. How do we use the quadratic formula to solve equations in general?
2. Why does $a^2 + b^2 = (a+b)^2 - 2ab$ work as an identity?
3. Could you calculate $a^4b + ab^4$ for the same $a$ and $b$?
4. How would the solution change if $ab = -7$?
5. What are the applications of symmetric polynomials like $a^3b + ab^3$?

Tip: Always double-check quadratic solutions by substituting back into the original equation!