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Question Number 212224 by RojaTaniya last updated on 07/Oct/24

a, b, c \in R

a + b + c = 1, a^{2} + b^{2} + c^{2} = 1

a^{10} + b^{10} + c^{10} = 1, a^{4} + b^{4} + c^{4} = ?

Answered by mr W last updated on 07/Oct/24

1^{2} = 1 + 2 (a b + b c + c a)

\Rightarrow a b + b c + c a = 0

0^{2} = a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} + 2 a b c (a + b + c)

\Rightarrow a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = - 2 a b c = - 2 k

{(- 2 k)}^{2} = a^{4} b^{4} + b^{4} c^{4} + c^{4} a^{4} + 2 a^{2} b^{2} c^{2} (a^{2} + b^{2} + c^{2})

\Rightarrow a^{4} b^{4} + b^{4} c^{4} + c^{4} a^{4} = 2 k^{2}

1^{2} = a^{4} + b^{4} + c^{4} + 2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})

\Rightarrow a^{4} + b^{4} + c^{4} = 1 + 4 k

a^{6} + b^{6} + c^{6} + (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}) (a^{2} + b^{2} + c^{2}) - 3 a^{2} b^{2} c^{2} = (1 + 4 k) \times 1

\Rightarrow a^{6} + b^{6} + c^{6} = 1 + 6 k + 3 k^{2}

{(1 + 4 k)}^{2} = a^{8} + b^{8} + c^{8} + 2 (a^{4} b^{4} + b^{4} c^{4} + c^{4} a^{4})

\Rightarrow a^{8} + b^{8} + c^{8} = 1 + 8 k + 12 k^{2}

a^{10} + b^{10} + c^{10} + (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}) (a^{6} + b^{6} + c^{6}) - a^{2} b^{2} c^{2} (a^{4} + b^{4} + c^{4}) = (1 + 8 k + 12 k^{2}) \times 1

1 + (- 2 k) (1 + 6 k + 3 k^{2}) - k^{2} (1 + 4 k) = (1 + 8 k + 12 k^{2}) \times 1

\Rightarrow k (2 k + 1) (k + 2) = 0

\Rightarrow k = 0, - \frac{1}{2}, - 2

\Rightarrow a^{4} + b^{4} + c^{4} = 1 + 4 k = 1, - 1, - 7

i f a, b, c \in R : a^{4} + b^{4} + c^{4} = 1

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