IIT JEE , AIEEE Forum - Mathematics , Algebra

shreyaarya

to an extent same methods as used above but i have just used a conditional identity which can be proved easily using the above methods of multiplication of

( a² + b² + c² ).( a³ + b³ + c³ )etc..

a + b + c = 0

ab + bc + ca = 3

abc = -3

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

implies a³ + b³ + c³ = -9(since a +b +c=0)

now,

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

0 = a² + b² + c² + 2(3)

Thus, a² + b² + c² = -6

here we can use the conditional identity

( a² + b² + c² )/2.( a³ + b³ + c³ )/3= a⁵ + b⁵ + c⁵/5

^{_{when a +b+c=0}}

now substituting values

we get -6/2.-9/3.5=a⁵ + b⁵ + c⁵

hence a⁵ + b⁵ + c⁵=-3.-3.5=45

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