All to the power 7?

Algebra Level 3

If α + β + γ = 4 \alpha + \beta + \gamma = 4 , α 2 + β 2 + γ 2 = 14 \alpha^2 + \beta^2 + \gamma^2 = 14 , and α 3 + β 3 + γ 3 = 34 \alpha^3 + \beta^3 + \gamma^3 = 34 , find α 7 + β 7 + γ 7 \alpha^7 + \beta^7 + \gamma^7 .


The answer is 2314.

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Animesh Sharma by animesh sharma , 20, India

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2 solutions

Chew-Seong Cheong
Jun 27, 2018

Relevant wiki: Newton's Identities

We can solve this problem using Newton's sums method (Newton's identities). First, let P n = α n + β n + γ n P^n = \alpha^n+\beta^n+\gamma^n , where n n is a positive integer. S 1 = α + β + γ S_1 = \alpha+\beta+\gamma , S 2 = α β + β γ + γ α S_2 = \alpha\beta+\beta\gamma+\gamma\alpha , and S 3 = α β γ S_3 = \alpha\beta\gamma . Then,

P 1 = S 1 = 4 P 2 = S 1 P 1 − 2 S 2 = 4 ( 4 ) − 2 S 1 = 14 ⟹ S 2 = 1 P 3 = S 1 P 2 − S 2 P 1 + 3 S 3 = 4 ( 14 ) − 4 + 3 S 3 = 34 ⟹ S 3 = − 6 P 4 = S 1 P 3 − S 2 P 2 + S 3 P 1 = 4 ( 34 ) − 14 − 6 ( 4 ) = 98 P 5 = S 1 P 4 − S 2 P 3 + S 3 P 2 = 4 ( 98 ) − 34 − 6 ( 14 ) = 274 P 6 = S 1 P 5 − S 2 P 4 + S 3 P 3 = 4 ( 274 ) − 98 − 6 ( 34 ) = 794 P 7 = S 1 P 6 − S 2 P 5 + S 3 P 4 = 4 ( 794 ) − 274 − 6 ( 98 ) = 2314 \begin{aligned} P_1 & = S_1 = 4 \\ P_2 & = S_1P_1-2S_2 = 4(4)-2S_1 = 14 & \small \color{#3D99F6} \implies S_2 = 1 \\ P_3 & = S_1P_2-S_2P_1+3S_3 = 4(14)-4+3S_3 = 34 & \small \color{#3D99F6} \implies S_3 = -6 \\ P_4 & = S_1P_3-S_2P_2+S_3P_1 = 4(34)-14 - 6(4) = 98 \\ P_5 & = S_1P_4-S_2P_3+S_3P_2 = 4(98)-34 - 6(14) = 274 \\ P_6 & = S_1P_5-S_2P_4+S_3P_3 = 4(274)-98 - 6(34) = 794 \\ P_7 & = S_1P_6-S_2P_5+S_3P_4 = 4(794)-274 - 6(98) = 2314 \end{aligned}

Therefore P 7 = α 7 + β 7 + γ 7 = 2314 P_7 = \alpha^7+\beta^7+\gamma^7 = \boxed{2314} .

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Zico Quintina
Jun 27, 2018

We re-write the equations:

α + β = 4 − γ ( 1 ) α 2 + β 2 = 14 − γ 2 ( 2 ) α 3 + β 3 = 34 − γ 3 ( 3 ) \begin{array}{rlll} \alpha + \beta &= \ \ 4 - \gamma & & \small (1) \\ \\ \alpha^2 + \beta^2 &= \ \ 14 - \gamma^2 & & \small (2) \\ \\ \alpha^3 + \beta^3 &= \ \ 34 - \gamma^3 & & \small (3) \end{array}

α 2 + 2 α β + β 2 = 16 − 8 γ + γ 2 ( 4 ) [ Square of ( 1 ) ] α β = γ 2 − 4 γ + 1 ( 5 ) [ Half the difference ( 4 ) − ( 2 ) ] α 2 − α β + β 2 = 13 + 4 γ − 2 γ 2 ( 6 ) [ ( 2 ) − ( 5 ) ] \begin{array}{rlllcccl} \alpha^2 + 2 \alpha \beta + \beta^2 &= \ \ 16 - 8\gamma + \gamma^2 & & \small (4) & & & & \small \text{[ Square of } (1) \ ] \\ \\ \alpha \beta &= \ \ \gamma^2 - 4 \gamma + 1 & & \small (5) & & & & \small \text{[ Half the difference } (4) - (2) \ ] \\ \\ \alpha ^2 - \alpha \beta + \beta^2 &= \ \ 13 + 4 \gamma - 2 \gamma^2 & & \small (6) & & & & \small [ \ (2) - (5) \ ] \\ \\ \end{array}

( α + β ) ( α 2 − α β + β 2 ) = 34 − γ 3 [ Factoring ( 3 ) ] ( 4 − γ ) ( 13 + 4 γ − 2 γ 2 ) = 34 − γ 3 [ Substituting ( 1 ) and ( 6 ) ] 52 + 3 γ − 12 γ 2 + 2 γ 3 = 34 − γ 3 3 γ 3 − 12 γ 2 + 3 γ + 18 = 0 3 ( γ − 3 ) ( γ − 2 ) ( γ + 1 ) = 0 γ = 3 , 2 , - 1 \begin{array}{rlllcccl} (\alpha + \beta) \ (\alpha^2 - \alpha \beta + \beta^2) &= \ \ 34 - \gamma^3 & & & & & & \small \text{[ Factoring } (3) \ ] \\ \\ (4 - \gamma) \ (13 + 4 \gamma - 2 \gamma^2) &= \ \ 34 - \gamma^3 & & & & & & \small \text{[ Substituting } (1) \text{ and } (6) \ ] \\ \\ 52 + 3 \gamma - 12 \gamma^2 + 2 \gamma^3 &= \ \ 34 - \gamma^3 \\ \\ 3 \gamma^3 - 12 \gamma^2 + 3 \gamma + 18 &= \ \ 0 \\ \\ 3 (\gamma - 3) \ (\gamma - 2) \ (\gamma + 1) &= \ \ 0 \\ \\ \gamma &= \ \ 3, 2, \text{-}1 \end{array}

Using these three values in our original equations, we quickly find that ( α , β , γ ) (\alpha, \beta, \gamma) can be any permutation of ( 3 , 2 , - 1 ) (3, 2, \text{-}1) .

Regardless of which permutation we use, we will get that α 7 + β 7 + γ 7 = 2314 \alpha^7 + \beta^7 + \gamma^7 = \boxed{2314}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Oh this is a very unusual way to find the values of α , β , γ \alpha,\beta, \gamma . I would have just used the algebraic identities :

a 2 + b 2 + c 2 = ( a + b + c ) 2 − 2 ( a b + a c + b c ) a^2 + b^2 +c^2 = (a+b+c)^2 - 2(ab+ ac + bc) and a 3 + b 3 + c 3 − 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 − ( a b + a c + b c ) ) a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - (ab+ac+bc)) .

Thanks for sharing!

Pi Han Goh - 2 years, 11 months ago

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Thanks for pointing these out; I don't think I've ever seen the second one. I did know the first one but my brain still thinks mostly in terms of two variables; gonna have to work on that!

zico quintina - 2 years, 11 months ago

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