IIT JEE , AIEEE Forum - Mathematics , Algebra

iberis22

1. a + b + c = 0

2. ab + bc + ca = 3

3. abc = -3

4. Since, a, b, c are the roots of equation,

a³ + 3a + 3 = 0

b³ + 3b + 3 = 0

c³ + 3c + 3 = 0

Thus, a³+ b³ + c³ + 3(a+b+c) + 9 = 0

But, a+b+c=0

Thus, a³ + b³ + c³ = -9

5. (a+b+c)² = a² + b² + c² + 2(ab + bc+ ca)

0 = a² + b² + c² + 2(3)

Thus, a² + b² + c² = -6

Now,

( a² + b² + c² )( a³ + b³ + c³ )

= a⁵ + b⁵ + c⁵ + a²b²(a+b) + b²c²(b+c) + a²c²(a+c)

putting a+b = -c

b+c = -a

c+a = -b

= a⁵ + b⁵ + c⁵ - abc(ab + bc + ca)

= a⁵ + b⁵ + c⁵ - (-3)(3)

LHS = (-6)(-9) = 54

Thus, a⁵ + b⁵ + c⁵ = 54-9 = 45