Roots To The 9th!

Algebra Level 3

If α , β , γ \alpha , \beta , \gamma are roots of the equation x 3 + 3 x + 9 = 0 , x^{3} + 3x + 9 = 0, find the value of α 9 + β 9 + γ 9 . \alpha^{9} + \beta^9 + \gamma^{9}.


The answer is 0.

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Ayush Parasar by Ayush Parasar , 22, India

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3 solutions

By Vieta's formulas we know that: α + β + γ = 0 \alpha+\beta+\gamma=0 , α β + α γ + β γ = 3 \alpha \beta+\alpha \gamma + \beta\gamma=3 and α β γ = − 9 \alpha \beta \gamma=-9 .

The fast way

Note that x 3 = − 3 ( x + 3 ) x^3=-3(x+3) , so x 9 = − 27 ( x 3 + 9 x 2 + 27 x + 27 ) x^9=-27(x^3+9x^2+27x+27) . Hence, the value we want is:

α 9 + β 9 + γ 9 = − 27 ( ( α 3 + β 3 + γ 3 ) + 9 ( α 2 + β 2 + γ 2 ) + 27 ( α + β + γ ) + 81 ) \alpha^9+\beta^9+\gamma^9=-27((\alpha^3+\beta^3+\gamma^3)+9(\alpha^2+\beta^2+\gamma^2)+27(\alpha+\beta+\gamma)+81)

α 2 + β 2 + γ 2 = ( α + β + γ ) 2 − 2 ( α β + α γ + β γ ) \alpha^2+\beta^2+\gamma^2=(\alpha + \beta + \gamma)^2-2(\alpha \beta+\alpha \gamma + \beta\gamma)

α 2 + β 2 + γ 2 = ( 0 ) 2 − 2 ( 3 ) = − 6 \alpha^2+\beta^2+\gamma^2=(0)^2-2(3)=-6

α 3 + β 3 + γ 3 = ( α + β + γ ) 3 − 3 ( α + β + γ ) ( α β + α γ + β γ ) + 3 ( α β γ ) \alpha^3+\beta^3+\gamma^3=(\alpha+\beta+\gamma)^3-3(\alpha+\beta+\gamma)(\alpha \beta+\alpha \gamma + \beta\gamma)+3(\alpha \beta \gamma)

α 3 + β 3 + γ 3 = ( 0 ) 3 − 3 ( 0 ) ( 3 ) + 3 ( − 9 ) = − 27 \alpha^3+\beta^3+\gamma^3=(0)^3-3(0)(3)+3(-9)=-27

α 9 + β 9 + γ 9 = − 27 ( − 27 + 9 ( − 6 ) + 27 ( 0 ) + 81 ) \alpha^9+\beta^9+\gamma^9=-27(-27+9(-6)+27(0)+81)

α 9 + β 9 + γ 9 = 0 \alpha^9+\beta^9+\gamma^9=\boxed{0}

The long way

By the identity:

α 3 + β 3 + γ 3 = ( α + β + γ ) 3 − 3 ( α + β + γ ) ( α β + α γ + β γ ) + 3 ( α β γ ) \alpha^3+\beta^3+\gamma^3=(\alpha+\beta+\gamma)^3-3(\alpha+\beta+\gamma)(\alpha \beta+\alpha \gamma + \beta\gamma)+3(\alpha \beta \gamma)

α 3 + β 3 + γ 3 = ( 0 ) 3 − 3 ( 0 ) ( 3 ) + 3 ( − 9 ) = − 27 \alpha^3+\beta^3+\gamma^3=(0)^3-3(0)(3)+3(-9)=-27

Similarly:

α 9 + β 9 + γ 9 = ( α 3 + β 3 + γ 3 ) 3 − 3 ( α 3 + β 3 + γ 3 ) ( ( α β ) 3 + ( α γ ) 3 + ( β γ ) 3 ) + 3 ( α β γ ) 3 \alpha^9+\beta^9+\gamma^9=(\alpha^3+\beta^3+\gamma^3)^3-3(\alpha^3+\beta^3+\gamma^3)((\alpha \beta)^3+(\alpha \gamma)^3+(\beta \gamma)^3)+3(\alpha \beta \gamma)^3

Again, in a similar way:

( α β ) 3 + ( α γ ) 3 + ( β γ ) 3 = ( α β + α γ + β γ ) 3 − 3 α β γ ( α β + α γ + β γ ) ( α + β + γ ) + 3 ( α β γ ) 2 (\alpha \beta)^3+(\alpha \gamma)^3+(\beta \gamma)^3=(\alpha\beta+\alpha\gamma+\beta\gamma)^3-3\alpha\beta\gamma(\alpha\beta+\alpha\gamma+\beta\gamma)(\alpha+\beta+\gamma)+3(\alpha\beta\gamma)^2

So,

α 9 + β 9 + γ 9 = ( α 3 + β 3 + γ 3 ) 3 − 3 ( α 3 + β 3 + γ 3 ) ( ( α β + α γ + β γ ) 3 − 3 α β γ ( α β + α γ + β γ ) ( α + β + γ ) + 3 ( α β γ ) 2 ) + 3 ( α β γ ) 3 \alpha^9+\beta^9+\gamma^9=(\alpha^3+\beta^3+\gamma^3)^3-3(\alpha^3+\beta^3+\gamma^3)((\alpha\beta+\alpha\gamma+\beta\gamma)^3-3\alpha\beta\gamma(\alpha\beta+\alpha\gamma+\beta\gamma)(\alpha+\beta+\gamma)+3(\alpha\beta\gamma)^2)+3(\alpha \beta \gamma)^3

Substituting the known values we arrive to:

α 9 + β 9 + γ 9 = ( − 27 ) 3 − 3 ( − 27 ) ( ( 3 ) 3 − 3 ( − 9 ) ( 6 ) ( 0 ) + 3 ( − 9 ) 2 ) + 3 ( − 9 ) 3 \alpha^9+\beta^9+\gamma^9=(-27)^3-3(-27)((3)^3-3(-9)(6)(0)+3(-9)^2)+3(-9)^3

α 9 + β 9 + γ 9 = 0 \alpha^9+\beta^9+\gamma^9=\boxed{0}

We can also do this by Newton's Sums, but since 3 2 = 9 3^2=9 , these are faster ways.

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Note: In the fast way, you could use α 3 = − 3 α − 9 \alpha^3 = - 3 \alpha - 9 , and hence ∑ α 3 = − 27 − 3 ∑ α = − 27 \sum \alpha^3 = -27 - 3 \sum \alpha = - 27 .

Calvin Lin Staff - 6 years, 5 months ago

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Another way to make it a bit shorter is to substitute x 3 x^{3} into the expansion of ( − 3 x − 9 ) 3 (-3x-9)^{3} to give -81( 3 x 2 x^{2} +8 x {x} +6). From there we calculate 3 x 2 x^{2} = α 2 \alpha^{2} + β 2 \beta^{2} + γ 2 \gamma^{2} = -6, 8 x {x} = 0, such that the expression = 0

Curtis Clement - 6 years, 5 months ago

Oh yes, a faster way :D

Alan Enrique Ontiveros Salazar - 6 years, 5 months ago

Much faster way , learnt in Adity raut's note - Bashing Unavailable

t n = t n − 1 + 1 2 t n − 2 + 1 6 t n − 3 t_{n} = t_{n-1} + \dfrac{1}{2}t_{n - 2} + \dfrac{1}{6}t_{n-3} ( t n = a n + b n + c n t_{n} = a^n + b^n + c^n )

Now α = a , β = b , γ = c \alpha =a , \beta=b , \gamma=c

B y V i e t a , a + b + c = 0 ~By ~Vieta~, a + b + c = 0

Applying the general term above ,

a 2 + b 2 + c 2 = 0 a^2 + b^2 + c^2 = 0 , this implies t 3 = 0 t_{3} = 0 and successive quadratics are dependent on the preceding , thus each is zero!

U Z - 6 years, 5 months ago

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@megh choksi You are referring to Newton's Identity, You have to generate the recurrence for each polynomial. The actual recurrence relation is:

The one that you stated is the recurrence relation for Aditya's example, and does not hold for this problem.

Note also that t 0 = 3 ≠ 0 t_0 = 3 \neq 0 . t 3 = − 27 ≠ 0 t_3 = -27 \neq 0 . If each of the powers have a sum of 0, then the variables must all be 0.

Calvin Lin Staff - 6 years, 5 months ago

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@Calvin Lin – Oh thanks , I did the same mistake in another problem too

U Z - 6 years, 5 months ago

writing ∑ α \sum \alpha means cyclic i.e α + β + γ = 0 \alpha + \beta + \gamma = 0 so it gives -27

U Z - 6 years, 5 months ago

Exactly what I did !! Look at my solution.

Akshat Sharda - 5 years, 6 months ago
Chew-Seong Cheong
Dec 26, 2014

The problem can be solved using Newton's Sums method.

Let S 1 = α + β + γ = 0 S_1 = \alpha + \beta + \gamma = 0 , S 2 = α β + β γ + γ α = 3 S_2 = \alpha\beta + \beta \gamma + \gamma\alpha = 3 and S 3 = α β γ = − 9 S_3 = \alpha\beta\gamma = -9 ; then P n = α n + β n + γ n P_n = \alpha^n + \beta^n + \gamma^n , where n = 1 , 2 , 3... n = 1,2,3... is given by:

{ P 1 = S 1 = 0 P 2 = S 1 P 1 − 2 S 2 = 0 − 2 ( 3 ) = − 6 P 3 = S 1 P 2 − S 2 P 1 + 3 S 3 = 0 − 0 + 3 ( − 9 ) = − 27 P 4 = S 1 P 3 − S 2 P 2 + S 3 P 1 = 0 − 3 ( − 6 ) + 0 = 18 P 5 = S 1 P 4 − S 2 P 3 + S 3 P 2 = 0 − 3 ( − 27 ) − 9 ( 6 ) = 135 P 6 = S 1 P 5 − S 2 P 4 + S 3 P 3 = 0 − 3 ( 18 ) − 9 ( − 27 ) = 189 P 7 = S 1 P 6 − S 2 P 5 + S 3 P 4 = 0 − 3 ( 135 ) − 9 ( 18 ) = − 567 P 8 = S 1 P 7 − S 2 P 6 + S 3 P 5 = 0 − 3 ( 189 ) − 9 ( 135 ) = − 1782 P 9 = S 1 P 8 − S 2 P 7 + S 3 P 6 = 0 − 3 ( − 567 ) − 9 ( 189 ) = 0 \begin {cases} P_1 = S_1 & & = 0 \\ P_2 = S_1 P_1 -2S_2 & = 0-2(3) & = -6 \\ P_3 = S_1 P_2-S_2P_1 + 3S_3 & = 0 - 0 +3(-9) & = -27 \\ P_4 = S_1P_3-S_2P_2 + S_3P_1 & = 0-3(-6)+0 & = 18 \\ P_5 = S_1P_4-S_2P_3 + S_3P_2 & = 0-3(-27)-9(6) & = 135 \\ P_6 = S_1P_5-S_2P_4 + S_3P_3 & = 0-3(18)-9(-27) & = 189 \\ P_7 = S_1P_6-S_2P_5 + S_3P_4 & = 0-3(135)-9(18) & = -567 \\ P_8 = S_1P_7-S_2P_6 + S_3P_5 & = 0-3(189)-9(135) & = -1782 \\ P_9 = S_1P_8-S_2P_7 + S_3P_6 & = 0-3(-567)-9(189) & = 0 \end {cases}

Therefore, P 9 = α 9 + β 9 + γ 9 = 0 P_9 = \alpha^9 + \beta^9 + \gamma^9 = \boxed{0} .

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Akshat Sharda
Nov 30, 2015

x 3 + 3 x + 9 = 0 ⇒ x 3 = − 3 ( x + 3 ) x 9 = − 27 ( x 3 + 9 x 2 + 27 x + 27 ) ∑ x 9 = − 27 ( ∑ ( x 3 + 9 x 2 + 27 x + 27 ) ) = − 27 ( ∑ x 3 + 9 ∑ x 2 + 27 ∑ x + ∑ 27 ) By Newton’s Sums : ∑ x = 0 , ∑ x 2 = − 6 and ∑ x 3 = − 27 = − 27 ( − 27 + 9 ( − 6 ) + 27 ( 0 ) + 27 × 3 ) = − 27 ( 0 ) = 0 x^3+3x+9=0\Rightarrow x^3=-3(x+3) \\ x^9=-27(x^3+9x^2+27x+27) \\ \sum x^9=-27\left(\sum(x^3+9x^2+27x+27)\right) \\ =-27\left(\sum x^3+9\sum x^2+27\sum x+\sum 27\right) \\ \text{By Newton's Sums : }\sum x=0,\sum x^2=-6 \text{ and } \sum x^3=-27 \\ =-27(-27+9(-6)+27(0)+27×3) \\ =-27(0)=\boxed{0}

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