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Algebra Level 4

a + b + c = 1 a 2 + b 2 + c 2 = 2 a 3 + b 3 + c 3 = 3 a 5 + b 5 + c 5 = ? \begin{array} { l l l l l l l l l } a & + & b & + & c & = & 1 \\ a^2 & + & b^2 & + & c^2 & = & 2 \\ a^3 & + & b^3 & + & c^3 & = & 3 \\ a^5 & + & b^5 & + & c^5 & = & ? \\ \end{array}


The answer is 6.

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Raghvendra Sonarikar by Raghvendra Sonarikar , 22, India

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2 solutions

Bhargav Pavuluri
Sep 1, 2015

5 Helpful 0 Interesting 0 Brilliant 1 Confused

Newton’s Sum is a much easier and shorter way. Let us consider a polynomial P ( x ) = x 3 + k x 2 + l x + m which have a,b,c as its roots and S n = a n + b n + c n By Newton’s Identities: ( 1 ) S 1 + k = 0 ⇒ k = − 1 ( 1 ) S 2 + ( k ) S 1 + 2 l = 0 ⇒ l = − 1 2 ( 1 ) S 3 + ( k ) S 2 + ( l ) S 1 + 3 m = 0 ⇒ m = − 1 6 Therefore, P ( x ) = x 3 − x 2 − 1 2 x − 1 6 ( 1 ) S 4 + ( − 1 ) S 3 + ( − 1 2 ) S 2 + ( − 1 6 ) S 1 = 0 ⇒ S 4 = 25 6 ( 1 ) S 5 + ( − 1 ) S 4 + ( − 1 2 ) S 3 + ( − 1 6 ) S 2 = 0 ⇒ S 5 = 6 \text{Newton's Sum is a much easier and shorter way.} \\ \text{Let us consider a polynomial } P(x)=x^{3}+kx^{2}+lx+m \text{ which have a,b,c as its roots and } S_{n}=a^{n}+b^{n}+c^{n} \\ \text{ By Newton's Identities:} \\ (1)S_{1}+k=0\Rightarrow k=-1 \\ (1)S_{2}+(k)S_{1}+2l=0\Rightarrow l=-\frac{1}{2} \\ (1)S_{3}+(k)S_{2}+(l)S_{1}+3m=0\Rightarrow m=-\frac{1}{6} \\ \text{Therefore, }P(x)=x^{3}-x^{2}-\frac{1}{2}x-\frac{1}{6} \\ (1)S_{4}+(-1)S_{3}+\left(-\frac{1}{2}\right)S_{2}+\left(-\frac{1}{6}\right)S_{1}=0\Rightarrow S_{4}=\frac{25}{6} \\ (1)S_{5}+(-1)S_{4}+\left(-\frac{1}{2}\right)S_{3}+\left(-\frac{1}{6}\right)S_{2}=0\Rightarrow S_{5}=\boxed{6}

Akshat Sharda - 5 years, 8 months ago

Ya i agree by newton sums it is only pf 3 min

Mohit Gupta - 5 years, 8 months ago
Devansh Shah
Nov 7, 2015

Multiply first and second equation and separate out the expression of who h you want to find out value on one side of equation. Now get value of cyclic summation of xy and value of xyz from first 2 equations to get answer as 6

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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