Looks simple right?

Algebra Level 3

a 5 + b 5 + c 5 a b + b c + c a \large \frac{ a^5 + b^5 + c^5 }{ ab + bc + ca }

Define non-zero values of a , b , c a,b,c such that their sum is 0. Which of these answer choices is equals to the expression above?

This problem is a part of the sets - 3's & 4's & Questions .
− 3 a b c -3abc − 1 -1 − 2 a b c -2abc − 5 a b c -5abc − a b c -abc a b c abc

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Sakanksha Deo by Sakanksha Deo , 23, India

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3 solutions

Sakanksha Deo
Mar 16, 2015

Again, my favourite, newton's sum,

Let,

S 1 = a + b + c S_1 = a + b + c

S 2 = a 2 + b 2 + c 2 S_2 = a^2 + b^2 + c^2

S 3 = a 3 + b 3 + c 3 S_3 = a^3 + b^3 + c^3

S 4 = a 4 + b 4 + c 4 S_4 = a^4 + b^4 + c^4

S 5 = a 5 + b 5 + c 5 S_5 = a^5 + b^5 + c^5

P 2 = a b + b c + a c P_2 = ab + bc + ac

P 3 = a b c P_3 = abc

Now,

According to newton's sum,

S 2 = S 1 2 − 2 P 2 ⇒ S 2 = − 2 P 2 S_2 = S_1^2 - 2 P_2 \Rightarrow S_2 = - 2 P_2

S 3 = S 1 S 2 − P 2 S 1 + 3 P 3 ⇒ S 3 = 3 P 3 S_3 = S_1 S_2 - P_2 S_1 + 3 P_3 \Rightarrow S_3 = 3 P_3

S 5 = S 4 S 1 − P 2 S 3 + P 3 S 2 ⇒ S 5 = 0 − 3 P 3 P 2 − 2 P 2 P 3 = − 5 P 3 P 2 ⇒ S 5 P 2 = − 5 P 3 S_5 = S_4 S_1 - P_2 S_3 + P_3 S_2 \Rightarrow S_5 = 0 - 3 P_3 P_2 - 2 P_2 P_3 = - 5 P_3 P_2 \Rightarrow \frac{ S_5 }{ P_2 } = - 5 P_3

⇒ a 5 + b 5 + c 5 a b + b c + a c = − 5 a b c \Rightarrow \frac{ a^5 + b^5 + c^5 }{ ab + bc + ac } = \boxed{ - 5abc }

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Yes, when we see the sum of the same powers ( a 5 + b 5 + c 5 ) a^5+b^5+c^5) , one should immediately consider Newton's Sum as a viable approach.

can you please elaborate the solution?

Aravindh D Schrösolver - 6 years, 1 month ago
Sujoy Roy
Mar 16, 2015

Put a = 2 , b = c = − 1 a=2, b=c=-1 , we get

a 5 + b 5 + c 5 a b + b c + c a \frac{a^5+b^5+c^5}{ab+bc+ca}

= 2 5 + ( − 1 ) 5 + ( − 1 ) 5 − 2 + 1 − 2 =\frac{2^5+(-1)^5+(-1)^5}{-2+1-2}

= − 10 = − 5 ∗ 2 ∗ ( − 1 ) ∗ ( − 1 ) = − 5 a b c =-10=-5*2*(-1)*(-1)=-5abc

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

This solution is wrong. You have only shown that it's true when a = 2 , b = c = − 1 a=2, b=c=-1 .

ohhh boy i too taken the same set of values...............

Madhukar Thalore - 6 years ago

If it is known beforehand that the relation is true for all values which satisfy the given conditions, is it not sufficient to prove it for only one set of values?

sujoy roy - 5 years, 11 months ago
Rahul Saxena
Apr 7, 2015

Just put any random values and you are there.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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