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Algebra Level 4

α 6 + β 6 + γ 6 + ( α β ) 3 + ( β γ ) 3 + ( α γ ) 3 \large\alpha^6 + \beta^6 + \gamma^6 + (\alpha\beta)^3 + (\beta\gamma)^3 + (\alpha\gamma)^3

Let α , β \alpha, \beta and γ \gamma be the zeroes of the polynomial x 3 − x − 1 x^3-x-1 . Find the value of the expression above.


The answer is 7.

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Puneet Pinku by Puneet Pinku , 20, India

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4 solutions

Chew-Seong Cheong
May 22, 2016

As α \alpha , β \beta and γ \gamma are zeroes or roots of x 3 − x − 1 = 0 x^3 - x - 1 = 0 , by Vieta's formulas, we have: { α + β + γ = 0 α β + β γ + γ α = − 1 \begin{cases} \color{#3D99F6}{\alpha + \beta + \gamma = 0} \\ \color{#D61F06}{\alpha \beta + \beta \gamma + \gamma \alpha = -1} \end{cases} .

From x 3 − x − 1 = 0 x^3-x-1 = 0 , ⟹ x 3 = x + 1 \implies x^3 = x+1

⟹ { α 3 = α + 1 ⟹ α 6 = ( α + 1 ) 2 = α 2 + 2 α + 1 β 3 = β + 1 ⟹ β 6 = ( β + 1 ) 2 = β 2 + 2 β + 1 γ 3 = γ + 1 ⟹ γ 6 = ( γ + 1 ) 2 = γ 2 + 2 γ + 1 \implies \begin{cases} \alpha^3 = \alpha + 1 & \implies \alpha^6 = (\alpha + 1)^2 = \alpha^2 + 2 \alpha + 1 \\ \beta^3 = \beta + 1 & \implies \beta^6 = (\beta + 1)^2 = \beta^2 + 2 \beta + 1 \\ \gamma^3 = \gamma + 1 & \implies \gamma^6 = (\gamma + 1)^2 = \gamma^2 + 2 \gamma + 1 \end{cases}

⟹ α 6 + β 6 + γ 6 = α 2 + β 2 + γ 2 + 2 ( α + β + γ ) + 3 = ( α + β + γ ) 2 − 2 ( α β + β γ + γ α ) + 2 ( α + β + γ ) + 3 = 0 − 2 ( − 1 ) + 2 ( 0 ) + 3 = 5 \begin{aligned} \implies \alpha^6 + \beta^6 + \gamma^6 & = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha + \beta + \gamma) + 3 \\ & = (\color{#3D99F6}{\alpha + \beta + \gamma})^2 - 2(\color{#D61F06}{\alpha \beta + \beta \gamma + \gamma \alpha}) + 2(\color{#3D99F6}{\alpha + \beta + \gamma}) + 3 \\ & = \color{#3D99F6}{0} - 2(\color{#D61F06}{-1}) + 2(\color{#3D99F6}{0}) + 3 \\ & = 5 \end{aligned}

⟹ { ( α β ) 3 = α 3 β 3 = ( α + 1 ) ( β + 1 ) = α β + α + β + 1 ( β γ ) 3 = β 3 γ 3 = ( β + 1 ) ( γ + 1 ) = β γ + β + γ + 1 ( γ α ) 3 = γ 3 α 3 = ( γ + 1 ) ( α + 1 ) = γ α + γ + α + 1 \implies \begin{cases} (\alpha\beta)^3 = \alpha^3 \beta^3 = (\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1 \\ (\beta\gamma)^3 = \beta^3 \gamma^3 = (\beta + 1)(\gamma + 1) = \beta\gamma + \beta + \gamma + 1 \\ (\gamma\alpha)^3 = \gamma^3 \alpha^3 = (\gamma+1)(\alpha + 1) = \gamma \alpha + \gamma + \alpha + 1 \end{cases}

⟹ ( α β ) 3 + ( β γ ) 3 + ( γ α ) 3 = α β + β γ + γ α + 2 ( α + β + γ ) + 3 = − 1 + 2 ( 0 ) + 3 = 2 \begin{aligned} \implies (\alpha\beta)^3 + (\beta\gamma)^3 + (\gamma\alpha)^3 & = \color{#D61F06}{\alpha \beta + \beta \gamma + \gamma \alpha} + 2(\color{#3D99F6}{\alpha + \beta + \gamma}) + 3 \\ & = \color{#D61F06}{-1} + 2(\color{#3D99F6}{0}) + 3 \\ & = 2 \end{aligned}

Therefore, α 6 + β 6 + γ 6 + ( α β ) 3 + ( β γ ) 3 + ( γ α ) 3 = 5 + 2 = 7 \alpha^6 + \beta^6 + \gamma^6 + (\alpha\beta)^3 + (\beta\gamma)^3 + (\gamma\alpha)^3 = 5 + 2 = \boxed{7}

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@Rishabh Cool why have you deleted your solution?

Rohit Udaiwal - 5 years ago

Brilliant method than mine!! But I couldn't completely understand the second part of the problem i.e. ( α β ) 3 (\alpha\beta)^3 part. How is it possible?? Please explain it to me.

Puneet Pinku - 5 years ago

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( α β ) 3 = α 3 β 3 = ( α + 1 ) ( β + 1 ) (\alpha\beta)^3 = \alpha^3 \beta^3 = (\alpha+1)(\beta+1)

Chew-Seong Cheong - 5 years ago

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Alright! Thankyou sir!!

Puneet Pinku - 5 years ago

Sir, could you also post a solution to my problem Disturbing Coefficients

Puneet Pinku - 5 years ago

Since we have α 3 = ( α + 1 ) \alpha^3=(\alpha+1) and β 3 = ( β + 1 ) \beta^3=(\beta+1) ,multiplying both equations we get ( α β ) 3 = ( α + 1 ) ( β + 1 ) . (\alpha\beta)^3=(\alpha+1)(\beta+1).

Rohit Udaiwal - 5 years ago

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Ah! Now I see that comes straight from the equation.initially I thought it was some kind of identity etc. Thankyou!

Puneet Pinku - 5 years ago
Puneet Pinku
May 21, 2016

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Aditya Dhawan
May 24, 2016

L e t f ( y ) b e a f u n c t i o n w h o s e r o o t s a r e c u b e s o f t h o s e o f f ( x ) = x 3 − x − 1 ⇒ y = x 3 ⟹ x = y 3 P l u g g i n g i n t o f ( x ) , w e g e t y − y 3 − 1 = 0 ⇒ y 3 − 3 y 2 + 2 y − 1 = 0 , w h o s e r o o t s a r e α 3 , β 3 , γ 3 N o w α 3 + β 3 + γ 3 = 3 a n d ( α β ) 3 + ( β γ ) 3 + ( α γ ) 3 = 2 ∴ α 6 + β 6 + γ 6 = ( α 3 + β 3 + γ 3 ) 2 − 2 [ ( α β ) 3 + ( β γ ) 3 + ( α γ ) 3 ] = 9 − 4 = 5 ∴ α 6 + β 6 + γ 6 + ( α β ) 3 + ( β γ ) 3 + ( α γ ) 3 = 7 Let\quad f(y)\quad be\quad a\quad function \quad whose\quad roots\quad are\quad cubes\quad of\quad those\quad of\quad { f(x)=x }^{ 3 }-x-1\\ \Rightarrow y={ x }^{ 3 }\Longrightarrow x=\sqrt [ 3 ]{ y } \\ \\ Plugging\quad into\quad f(x),\quad we\quad get\quad y-\sqrt [ 3 ]{ y } -1=0\\ \Rightarrow { y }^{ 3 }-3{ y }^{ 2 }+2y-1=0\quad ,whose\quad roots\quad are\quad { \alpha }^{ 3 }{ ,\beta }^{ 3 },{ \gamma }^{ 3 }\\ \\ Now\quad { \alpha }^{ 3 }{ +\beta }^{ 3 }+{ \gamma }^{ 3 }=3\quad and\quad { (\alpha \beta })^{ 3 }{ +(\beta \gamma ) }^{ 3 }+{ (\alpha \gamma ) }^{ 3 }=2\quad \\ \therefore \quad { \alpha }^{ 6 }+{ \beta }^{ 6 }+{ \gamma }^{ 6 }={ ({ \alpha }^{ 3 }{ +\beta }^{ 3 }+{ \gamma }^{ 3 }) }^{ 2 }-2\left[ { (\alpha \beta })^{ 3 }{ +(\beta \gamma ) }^{ 3 }+{ (\alpha \gamma ) }^{ 3 } \right] =\quad 9-4=5\\ \therefore { \quad \alpha }^{ 6 }+{ \beta }^{ 6 }+{ \gamma }^{ 6 }+{ (\alpha \beta })^{ 3 }{ +(\beta \gamma ) }^{ 3 }+{ (\alpha \gamma ) }^{ 3 }=\boxed { 7 } \\ \\

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Moderator note:

Good approach.

However, be careful with the variable substitution when it isn't bijective.

Awesome solution but why is the y function quadratic not cubic.

Puneet Pinku - 5 years ago

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It's a silly mistake. It was actually supposed to be a cubic function ( since f(x) is cubic). I have hence corrected it. Thanks! :)

Aditya Dhawan - 5 years ago

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I think you have missed -1 in f(x) in the firstline itself.

Puneet Pinku - 5 years ago

Very nice solution. Up voted.

Niranjan Khanderia - 2 years, 11 months ago

S t a n d a r d a p p r o a c h : V i e t a ′ s f o r m u l a s . { α + β + γ = 0 α β + β γ + γ α = − 1 α ∗ β ∗ γ = 1 X 2 + Y 2 + Z 2 = ( X + Y + Z ) 2 − 2 ∗ ( X Y + Y Z + Z X ) . . . . . . . . . . ( A ) X 3 + Y 3 + Z 3 = ( X + Y + Z ) ∗ { ( X + Y + Z ) 2 − 3 ∗ ( X Y + Y Z + Z X ) } + 3 ∗ X Y Z . . . . . . . . . . . ( B ) ∴ B y ( A ) α 2 + β 2 + γ 2 = ( α + β + γ ) 2 − 2 ∗ ( α β + β γ + γ α ) = 0 − 2 ∗ ( − 1 ) = 2. α 2 β 2 + β 2 γ 2 + γ 2 α 2 = ( α β + β γ + γ α ) 2 − 2 ∗ ( α ∗ β ∗ γ 2 + α ∗ β 2 ∗ γ + α 2 ∗ β ∗ γ ) = ( α β + β γ + γ α ) 2 − 2 ∗ α ∗ β ∗ γ ∗ ( α + β + γ ) = 1 − 2 ∗ 1 ∗ 0 = 1. ∴ B y ( B ) α 6 + β 6 + γ 6 = ( α 2 + β 2 + γ 2 ) ∗ { ( α 2 + β 2 + γ 2 ) 2 − 3 ∗ ( α 2 ∗ β 2 + β 2 ∗ γ 2 + γ 2 ∗ α 2 ) } + 3 ∗ ( α ∗ β ∗ γ ) 2 = 2 { 4 − 3 ∗ 1 } + 3 ∗ 1 = 5. ( α β ) 3 + ( β γ ) 3 + ( γ α ) 3 = ( α β + β γ + γ α ) { ( α β + β γ + γ α ) 2 − 3 ∗ ( α ∗ β ∗ γ 2 + α ∗ β 2 ∗ γ + α 2 ∗ β ∗ γ ) } + 3 ∗ ( α ∗ β ∗ γ ) 2 = ( α β + β γ + γ α ) { ( α β + β γ + γ α ) 2 − 3 ∗ α ∗ β ∗ γ ∗ ( α + β + γ ) } + 3 ∗ ( α ∗ β ∗ γ ) 2 = − 1 ( 1 − 0 ) + 3 = 2. ∴ α 6 + β 6 + γ 6 + ( α β ) 3 + ( β γ ) 3 + ( γ α ) 3 = 5 + 2 = 7. Standard ~approach:~~Vieta's~ formulas.\\ \begin{cases} \color{#3D99F6}{\alpha + \beta + \gamma = 0} \\ \color{#D61F06}{\alpha \beta + \beta \gamma + \gamma \alpha = -1}\\ \color{#BA33D6}{\alpha* \beta * \gamma = 1}\\ \end{cases}\\ \begin{aligned}\\ X^2+Y^2+Z^2 &=(X+Y+Z)^2-2*(XY+YZ+ZX)..........(A)\\ X^3+Y^3+Z^3 &=(X+Y+Z)* \Big \{(X+Y+Z)^2-3*(XY+YZ+ZX) \Big \} +3*XYZ...........(B)\\ ~~~\\~~~~\\ \therefore~By~(A)~~~~~~~\\ ~~~\\ \alpha^2 + \beta^2 + \gamma^2 &=( \alpha + \beta + \gamma )^2 - 2*(\alpha \beta + \beta \gamma + \gamma \alpha)\\ &=0-2*(-1)=2.\\ ~~~~\\ \alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2 &=(\alpha \beta + \beta \gamma + \gamma \alpha)^2 - 2*(\alpha* \beta * \gamma^2+\alpha* \beta^2 * \gamma+\alpha^2* \beta * \gamma)\\ &=(\alpha \beta + \beta \gamma + \gamma \alpha)^2-2*\alpha* \beta * \gamma*(\alpha + \beta + \gamma)\\ &=1-2*1*0=1.\\ ~~~~\\ ~~~~~\\ \therefore~By~(B)~~~~~~~\\ ~~~\\ \alpha^6 + \beta^6 + \gamma^6 &=( \alpha^2 + \beta^2 + \gamma^2 )* \Big \{( \alpha^2 + \beta^2 + \gamma^2 )^2 - 3*(\alpha^2* \beta^2+\beta^2 *\gamma^2 +\gamma^2*\alpha^2) \Big \}+ 3*(\alpha*\beta*\gamma)^2\\ &=2\{4-3*1\}+3*1=5.\\ ~~~~~~\\ (\alpha \beta)^3 + (\beta \gamma)^3 + (\gamma \alpha)^3 &=(\alpha \beta + \beta \gamma + \gamma \alpha) \Big \{(\alpha \beta + \beta \gamma + \gamma \alpha)^2-3*(\alpha* \beta * \gamma^2+\alpha* \beta^2 * \gamma+\alpha^2* \beta * \gamma) \Big \}+3*(\alpha*\beta*\gamma)^2\\ &=(\alpha \beta + \beta \gamma + \gamma \alpha) \Big \{(\alpha \beta + \beta \gamma + \gamma \alpha)^2-3*\alpha* \beta * \gamma*(\alpha + \beta + \gamma) \Big \}+3*(\alpha*\beta*\gamma)^2\\ &=-1(1-0)+3=2.\\ ~~~~\\~~~~\\ \therefore~~ \alpha^6 + \beta^6 + \gamma^6 + (\alpha \beta)^3 + (\beta \gamma)^3 + (\gamma \alpha)^3 &=5+2 =\Large \color{#D61F06}{7}.\\ \end{aligned}

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