If equation ax^2 + bx + c = 0 has roots Alpha and Beta, find values of these expressions in terms of a, b, and c. (1) alpha^4 + beta^4 (2) alpha^2/beta + beta^2/alpha (3) alpha^3/beta + beta^3/alpha (4) 1/(a*alpha + b) + 1/(a*beta + b)

Given that equation ax^2 + bx + c = 0 has roots α and β

=> α + β = -b/a

and αβ = c/a

Keeping these in mind, we'll now solve the problems.

1) α^4 + β^4

=> (α²)² + (β²)²

=> (α² + β²)² - 2α²β²        by using a² + b² = (a + b)² - 2ab

=> ((α + β)²  - 2αβ)² - 2(αβ)²        using the same identity again

We will now substitute the values of α + β and αβ we found in the beginning of the solution

=> ((-b/a)² - 2c/a)² - 2(c/a)²

=> (b²/a² - 2c/a)² - 2c²/a²

=> ((b² - 2ac)/a²)² - 2c²/a²

=> (b² - 2ac)²/a⁴ - 2c²/a²

=> ((b² - 2ac)² - 2a²c²) / a⁴
=> (b⁴ - 4ab²c + 4a²c² - 2a²c²) / a⁴

=> (b⁴ - 4ab²c + 2a²c²) / a⁴

2) α^2/β + β^2/α

=> (α³ + β³)/αβ

=> (α + β)(α² + β² - αβ) /αβ

Again, using a² + b² = (a + b)² - 2ab

=> (α + β)((α + β)² - 3αβ) / αβ

=> (-b/a)((-b/a)² - 3(c/a)) / (c/a)

=> (-b/a)(b²/a² - 3c/a) / (c/a)

=> (-b/c)((b² - 3ac)/ a²)

=> (-b³ + 3abc) / a²c 

3) α^3/β + β^3/α

=> (α⁴ + β⁴)/αβ

We've already calculated for (α⁴ + β⁴) in the first question. We will use the result here.

=> ((b⁴ - 4ab²c + 2a²c²) / a⁴) / αβ

=> ((b⁴ - 4ab²c + 2a²c²) / a⁴) / (c/a)

=> (b⁴ - 4ab²c + 2a²c²) / a³c

4) 1/(aα + b) + 1/(aβ + b)

We already know that 

α + β = -b/a

=> a(α + β) = -b

=> -a(α + β) = b

Substituting this value of b in the problem,

=> 1/(aα + b) + 1/(aβ + b)

=> 1/(aα - a(α + β)) + 1/(aβ - a(α + β))

=> 1/(-aβ) + 1/(-aα)

=> -1/a (1/β + 1/α)

=> -1/a (α + β)/αβ

=> -1/a (-b/a)(c/a)

=> b/ac