If alpha, beta are the roots of ax²+bx+c=0, find the values of:

i. (aplha/beta - beta/alpha)²

ii. alpha³/beta + beta³/alpha

Since let alpha =A beta = B. Then since A B are the roots of the equation therefore

Sum of roots = -b/a product of roots = c/a

= A + B = -b/a AB = c/a .........(1)

First Part : (A/B - B/A)²

(A/B - B/A)² = [(A² - B²)/(AB)]² = [(A + B)(A - B)/AB]²

= (A + B)²(A - B)²/(AB)² .........(2)

Also (A - B)² = (A + B)² - 4AB

= (A - B)² = (-b/a)² - 4(c/a) (Putting values from equ. (1) )

= (A - B)² = (b²/a²) - 4(c/a) = (b² - 4ac)/a² .........(3)

Now putting values from equ. (1) (3) in equ. (2) we get

(A/B - B/A)² = (-b/a)²{(b² - 4ac)/a²}/(c/a)² = (b²/a²){(b² - 4ac)/a²}/(c²/a²)

= (b²/a²){(b² - 4ac)/a²}(a²/c²)

= b²(b² - 4ac)/a²c².

Second Part: (A³/B + B³/A)

= (A⁴ + B⁴)/AB ............(4)

Now A⁴ + B⁴ = (A² + B²)² - 2A²B² = (A² + B²)² - 2(AB)²

= [ (A + B)² - 2AB]² - 2(AB)² ( since (A² + B²) = (A + B)² - 2AB)

Now putting values from eqn.(1) we get

[ (-b/a)² - 2(c/a)]² - 2(c/a)² = (b²/a² - 2c/a)² - 2(c/a)²

= (b²/a² - 2c/a + 2c/a)(b²/a² - 2c/a - 2c/a)

= (b²/a²)(b²/a² - 4c/a) = (b²/a²){(b² - 4ac)/a²} ..........(5)

Putting values from eqn. (5) (1) in eqn.(4) we get

(A⁴ + B⁴)/AB = (b²/a²) {(b² - 4ac)/a²}/(c/a)

= (b²/a²) {(b² - 4ac)/a²} (a/c)

= b²(b² - 4ac)/a³c