Let's go through each of the problems step by step:

a) $a^2 - 5a - 6 = 0$

This is a quadratic equation, which we can factor or solve using the quadratic formula.

To factor: $a^2 - 5a - 6 = (a - 6)(a + 1) = 0$

Thus, the solutions are: $a = 6 \quad \text{or} \quad a = -1$

b) $b^4 - 16 = 0$

This is a difference of squares.

We can write it as: $b^4 - 16 = (b^2 - 4)(b^2 + 4) = 0$

For the first factor: $b^2 - 4 = (b - 2)(b + 2) = 0 \quad \Rightarrow \quad b = 2 \quad \text{or} \quad b = -2$

For the second factor: $b^2 + 4 = 0 \quad \Rightarrow \quad b^2 = -4 \quad \Rightarrow \quad b = \pm 2i$

Thus, the solutions are: $b = 2, \, b = -2, \, b = 2i, \, b = -2i$

c) $49 - 14c + c^2 = 0$

We can rearrange this as a standard quadratic form: $c^2 - 14c + 49 = 0$

This is a perfect square: $(c - 7)^2 = 0$

Thus, the only solution is: $c = 7$

Would you like more details on any of the solutions? Here are some related questions you could explore:

1. What other methods can be used to solve quadratic equations like the one in part (a)?
2. How can we recognize a difference of squares, like in part (b)?
3. Could we use the quadratic formula for part (c)? How would that compare to factoring?
4. How do imaginary numbers factor into solving higher-degree polynomials?
5. What if the coefficient of $c^2$ in part (c) was different—how would the factoring change?

Tip: When solving quadratic equations, always check if factoring is possible before applying more complex methods like the quadratic formula.