If-a-b-c-1-a-2-b-2-c-2-2-a-3-b-3-c-3-3-then-a-5-b-5-c-5-

Question Number 57791 by Tawa1 last updated on 12/Apr/19

If a + b + c = 1

a^{2} + b^{2} + c^{2} = 2

a^{3} + b^{3} + c^{3} = 3

then a^{5} + b^{5} + c^{5} = ?

Answered by naka3546 last updated on 12/Apr/19

6

Commented by Tawa1 last updated on 12/Apr/19

Please workings

Answered by MJS last updated on 12/Apr/19

\det [\begin{matrix} a & b & c & 1 \\ a^{2} & b^{2} & c^{2} & 2 \\ a^{3} & b^{3} & c^{3} & 3 \\ a^{5} & b^{5} & c^{5} & x \end{matrix}] = 0

\Rightarrow x = a^{2} b c + {a b}^{2} c + {a b c}^{2} - 4 a b c - 2 a^{2} b - 2 a^{2} c - 2 b^{2} c - 2 {a b}^{2} - 2 {a c}^{2} - 2 {b c}^{2} + 3 a^{2} + 3 b^{2} + 3 c^{2 w} + 3 a b + 3 a c + 3 b c

(1) c = 1 - a - b

(2) b = \frac{- a + 1 + \sqrt{- 3 a^{2} + 2 a + 3}}{2}

\Rightarrow x = 3 a^{3} - 3 a^{2} - \frac{3}{2} a + \frac{11}{2}

(3) 3 a^{3} - 3 a^{2} - \frac{3}{2} a - \frac{1}{2} = 0

\Rightarrow x = 6

Commented by Tawa1 last updated on 12/Apr/19

God bless you sir

Commented by MJS last updated on 12/Apr/19

there should be an easier method but I cannot

remember .

anyway this is better than solving the 3^{rd}

degree polynome

Commented by Tawa1 last updated on 12/Apr/19

God bless you sir

Answered by naka3546 last updated on 12/Apr/19

Commented by Tawa1 last updated on 12/Apr/19

God bless you sir . But image too small

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