An Easy Version

Algebra Level 4

If the roots of p ( x ) = x 3 + 3 x 2 + 4 x − 8 p(x) = x^3 + 3x^2 + 4x - 8 are a a , b b , and c c , then find the value of a 2 ( 1 + a ) + b 2 ( 1 + b ) + c 2 ( 1 + c ) a^2 (1 + a) + b^2 (1 + b) + c^2 (1 + c)

Inspiration


The answer is 34.

Akhil Bansal by Akhil Bansal , 22, India

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3 solutions

Akshat Sharda
Dec 31, 2015

Let P n = a n + b n + c n P_{n}=a^n+b^n+c^n .

By using Newton's sums :

P 1 + 3 = 0 ⇒ P 1 = − 3 P 2 + ( 3 ) P 1 + 8 = 0 ⇒ P 2 = 1 P 3 + ( 3 ) P 2 + ( 4 ) P 1 − 24 = 0 ⇒ P 3 = 33 ∑ cyc a 2 ( a + 1 ) = P 2 + P 3 = 1 + 33 = 34 \begin{aligned} P_{1}+3 & =0\Rightarrow P_{1}=-3 \\ P_{2}+(3)P_{1}+8 & =0\Rightarrow P_{2}=1 \\ P_{3}+(3)P_{2}+(4)P_{1}-24 &=0 \Rightarrow P_{3}=33 \\ \displaystyle \sum_{\text{cyc}}a^2(a+1) & = P_{2}+P_{3} \\ & = 1+33 \\ & = \boxed{34} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

B y V i e t a , a + b + c = − 3 . . . . . a b + b c + c a = 4 . . . . . . a b c = 8. a 2 + b 2 + c 2 = ( a + b + c ) 2 − 2 ( a b + b c + c a ) = 9 − 8 = 1. a 3 + b 3 + c 3 = 3 a b c + ( a + b + c ) { a 2 + b 2 + c 2 − ( a b + b c + c a ) } = 3 ∗ 8 + ( − 3 ) ( 1 − 4 ) = 24 + 9 = 33. G i v e n E x p . = a 3 + b 3 + c 3 + a 2 + b 2 + c 2 = 33 + 1 = 34 By ~ Vieta,~ a+b+c= - 3 ~~..... ab+bc+ca=4~~......abc =8.\\ a^2+b^2+c^2=(a+b+c)^2 -2(ab+bc+ca)=9 - 8=1.\\ a^3+b^3+c^3=3abc+(a+b+c)\{a^2+b^2+c^2 - (ab+bc+ca) \}=3*8+(- 3)(1 - 4)=24+9=33.\\ Given ~Exp. =a^3+b^3+c^3~~+~~a^2+b^2+c^2=33+1= \Large \color{#D61F06}{34}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Yuri Lombardo
Nov 6, 2015

x 3 + 3 x 2 + 4 x − 8 x^{3}+3x^{2}+4x-8 can be written using Ruffini as: ( x − 1 ) ( x 2 + 4 x + 8 ) (x-1)(x^2+4x+8)

x 2 + 4 x + 8 x^{2}+4x+8 using quadratic formula we get: x = − 2 + − − 4 = − 2 + − 2 i x=-2+-\sqrt{-4}=-2+-2i

i i is the imaginary number: − 1 \sqrt{-1}

Than we get: a = 1 ; b = − 2 − 2 i ; c = − 2 + 2 i a = 1 ; b = -2-2i ; c = -2+2i And solving the equation becomes easy.

( a 2 ) ( a + 1 ) = 1 2 ) ( 1 + 1 ) = 2 (a^{2})(a+1) = 1^{2})(1+1) = 2

( b 2 ) ( b + 1 ) = ( − 2 − 2 i ) 2 ( − 2 − 2 i + 1 ) = ( 4 − 4 + 8 i ) ( − 1 − 2 i ) = 8 i ( − 1 − 2 i ) = 16 − 8 i (b^{2})(b+1) = (-2-2i)^{2}(-2-2i+1) = (4-4+8i)(-1-2i) = 8i(-1-2i) = 16-8i

( c 2 ) ( c + 1 ) = ( − 2 + 2 i ) 2 ( − 2 + 2 i + 1 ) = ( 4 − 4 − 8 i ) ( − 1 + 2 i ) = − 8 i ( − 1 + 2 i ) = 16 + 8 i (c^{2})(c+1) = (-2+2i)^{2}(-2+2i+1) = (4-4-8i)(-1+2i) = -8i(-1+2i) = 16+8i

In the end we got:

2 + 16 − 8 i + 16 + 8 i = 2 + 16 + 16 = 34 2+16-8i+16+8i = 2+16+16 = \boxed{34}

0 Helpful 0 Interesting 0 Brilliant 1 Confused

@Yuri Lombardo Wrap the math in \ ( math \ ) (without the spaces) or \ [ math \ ] if you want to display in the center.I've edited the first line of your solution,so that you can hover your mouse over it and see the changes that I've made so that you can make them in the solution yourself :).

Abdur Rehman Zahid - 5 years, 7 months ago

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Thank you very much, now i'll edit it all =D I followed the formatting guide correctly but i forgot to wrap math into the \parenthesis

Yuri Lombardo - 5 years, 7 months ago

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