If { ((B+R+P=−1)),((B^2 +R^2 +P^2 =17)),((B^3 +R^3 +P^3 =11)) :} then B^5 +R^5 +P^5 =?


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Question Number 186929 by cortano12 last updated on 12/Feb/23
$If { ((B+R+P=−1)),((B^2 +R^2 +P^2 =17)),((B^3 +R^3 +P^3 =11)) :} then B^5 +R^5 +P^5 =?$
$I f {\begin{cases} B + R + P = - 1 \\ B^{2} + R^{2} + P^{2} = 17 \\ B^{3} + R^{3} + P^{3} = 11 \end{cases}$ $t h e n B^{5} + R^{5} + P^{5} = ?$
	Answered by mr W last updated on 12/Feb/23

	$m e t h o d I$ $p_{1} = e_{1} = - 1$ $p_{2} = e_{1} p_{1} - 2 e_{2}$ $17 = {(- 1)}^{2} - 2 e_{2} \Rightarrow e_{2} = - 8$ $p_{3} = e_{1} p_{2} - e_{2} p_{1} + 3 e_{3}$ $11 = - 1 \times 17 - (- 8) (- 1) + 3 e_{3} \Rightarrow e_{3} = 12$ $p_{4} = e_{1} p_{3} - e_{2} p_{2} + e_{3} p_{1} = - 1 \times 11 + 8 \times 17 - 12 \times 1 = 113$ $p_{5} = e_{1} p_{4} - e_{2} p_{3} + e_{3} p_{2} = - 1 \times 113 + 8 \times 11 + 12 \times 17 = 179$ $i . e . B^{5} + R^{5} + P^{5} = 179$
	Answered by mr W last updated on 12/Feb/23

	$m e t h o d I I$ $i w r i t e a, b, c i n s t e a d o f B, R, P .$ $a + b + c = - 1$ ${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$ ${(- 1)}^{2} = 17 + 2 (a b + b c + c a)$ $\Rightarrow a b + b c + c a = - 8$ ${(a + b + c)}^{3} = a^{3} + b^{3} + c^{3} - 3 a b c + 3 (a + b + c) (a b + b c + c a)$ ${(- 1)}^{3} = 11 - 3 a b c + 3 (- 1) (- 8)$ $\Rightarrow a b c = 12$ ${(a b + b c + c a)}^{2} = a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} + 2 a b c (a + b + c)$ ${(- 8)}^{2} = a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} + 2 \times 12 (- 1)$ $\Rightarrow a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = 88$ $(a^{2} + b^{2} + c^{2}) (a^{3} + b^{3} + c^{3}) = a^{5} + b^{5} + c^{5} + (a + b + c) (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}) - a b c (a b + b c + c a)$ $(17) (11) = a^{5} + b^{5} + c^{5} + (- 1) (88) - 12 (- 8)$ $\Rightarrow a^{5} + b^{5} + c^{5} = 179$
	Answered by horsebrand11 last updated on 12/Feb/23

	$T_{n} = B^{n} + R^{n} + P^{n}$ $T_{0} = B^{0} + R^{0} + P^{0} = 3$ $T_{1} = - 1 = α_{1}$ $T_{2} = B^{2} + R^{2} + P^{2} = {(B + R + P)}^{2} - 2 (B R + B P + R P)$ $\Rightarrow 17 = 1 - 2 (B R + B P + R P)$ $\Rightarrow B R + B P + R P = - 8 = α_{2}$ $T_{n} = α_{1} T_{n - 1} - α_{2} T_{n - 2} + α_{3} T_{n - 3}$ $T_{3} = (- 1) . 17 - (- 8) (- 1) + α_{3} (3) = 11$ $\Rightarrow α_{3} = 12$ $T_{4} = α_{1} T_{3} - α_{2} T_{2} + α_{3} T_{1}$ $\Rightarrow T_{4} = (- 1) (11) - (- 8) (17) + (12) . (- 1)$ $\Rightarrow T_{4} = 113$ $T_{5} = α_{1} T_{4} - α_{2} T_{3} + α_{3} T_{2}$ $\Rightarrow T_{5} = (- 1) (113) - (- 8) (11) + (12) (17)$ $\Rightarrow T_{5} = 179$
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