Mathematical Enigma - XII

Algebra Level 4

Find the value of α 5 + β 5 + γ 5 { \alpha }^{ 5 }+{ \beta }^{ 5 }+{ \gamma }^{ 5 } , where α , β , γ \alpha ,\beta ,\gamma are the roots of x 3 + 3 x + 3 = 0 { x }^{ 3 }+3x+3=0 .


The answer is 45.

View wiki
Swapnil Das by Swapnil Das , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Jagdish Singh
Nov 2, 2015

Given α , β , γ \alpha,\beta,\gamma are the roots of the equation x 3 + 3 x + 3 = 0 x^3+3x+3=0 . So using Vieta theorem, we get

α + β + γ = 0 , α β + β γ + γ α = 3 , α β γ = − 3 \alpha+\beta+\gamma = 0\;\;, \alpha\beta+\beta\gamma+\gamma\alpha = 3\;\;,\alpha\beta\gamma = -3

Now If α + β + γ = 0 \alpha+\beta+\gamma = 0 Then α 3 + β 2 + γ 3 = 3 α β γ = − 9 ⇒ ∑ α 3 = − 9 \alpha^3+\beta^2+\gamma^3 = 3\alpha\beta\gamma = -9\Rightarrow \displaystyle \sum \alpha^3 = -9

and α 2 + β 2 + γ 2 = ∑ α 2 = ( α + β + γ ) 2 − 2 ( α β + β γ + γ α ) = − 6 \alpha^2+\beta^2+\gamma^2 = \displaystyle \sum \alpha^2 = (\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma \alpha) = -6

Given x 3 + 3 x + 3 = 0 × x 2 ⇒ x 5 + 3 x 3 + 3 x 2 = 0 x^3+3x+3 = 0\times x^2\Rightarrow x^5+3x^3+3x^2=0 . Now given α , β , γ \alpha,\beta,\gamma are the roots of that equation

So we get ∑ α 5 + 3 α 3 + ∑ α 2 = 0 ⇒ ∑ α 5 = 45 \displaystyle \sum \alpha^5+3\alpha^3+\sum \alpha^2 = 0\Rightarrow \displaystyle \sum \alpha^5 =45

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution. Typo in last line. It should be ∑ α 5 + 3 ∑ α 3 + 3 ∑ α 2 = 0 \sum\alpha ^5+3\sum\alpha ^3+3\sum \alpha ^2=0

Harish Sasikumar - 5 years, 7 months ago

Newton's sum !! It is way too easy !!

Akshat Sharda - 5 years, 6 months ago
Akhilesh Vibhute
Dec 11, 2015

S5+3S3+3S2=0 S2=-6 S3+3S1+9=0 Substitute and simplify life becomes absolutely easier............:-)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Just rewriting with Latex and in detail, so that it is easy to read. M u l t i p l y i n g t h e e q u a t i o n b y X 2 , X 5 + 3 X 3 + 3 X 2 = 0. ⟹ ∑ c y c l i c α 5 + 3 ∑ c y c l i c α 3 + 3 ∑ c y c l i c α 2 = 0. Using Vieta, and algebraic manipulation, ∑ c y c l i c α = 0 ∑ c y c l i c α 2 = − 6 , B u t ∑ c y c l i c α 3 + 3 ∗ ∑ c y c l i c α + ∑ c y c l i c 3 = 0. ⟹ ∑ c y c l i c α 3 + 0 + 9 = 0. ⟹ ∑ c y c l i c α 3 = − 9 . ∴ ∑ c y c l i c α 5 = 3 ∗ 9 + 3 ∗ 6 = 45 \text{Just rewriting with Latex and in detail, so that it is easy to read.}\\ Multiplying ~ the ~ equation ~ by ~ X^2, ~~~~~~~~~~X^5+3X^3+3X^2=0.\\ \implies ~\sum_{cyclic}\alpha ^5+ \color{#3D99F6}{3\sum_{cyclic}\alpha ^3+3\sum_{cyclic} \alpha ^2}=0.\\ \text{Using Vieta, and algebraic manipulation,} \\ \color{#3D99F6}{\sum_{cyclic}\alpha=0 ~~~~\sum_{cyclic}\alpha^2=- 6}, \\ But ~~\sum_{cyclic}\alpha ^3 +3*\sum_{cyclic}\alpha +\sum_{cyclic}3=0.\\ \implies ~ \sum_{cyclic}\alpha ^3+ 0 + 9=0. ~~~\implies \color{#3D99F6}{ \sum_{cyclic}\alpha ^3= -9}.\\ \therefore ~\sum_{cyclic}\alpha ^5=3*9+3*6= \Large ~~~\color{#D61F06}{45}

Niranjan Khanderia - 5 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...