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Algebra Level 4

If a a is a root of x 2 3 x + 1 x^2-3x+1 .Then If the value of a 3 a 6 + 1 \dfrac{a^3}{a^6 + 1} can be expressed as α β \dfrac{\alpha}{\beta} .Find α + β \alpha+\beta


The answer is 19.

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6 solutions

Vikram Waradpande
Dec 24, 2014

a 6 + 1 a 3 = a 3 + 1 a 3 = ( a + 1 a ) 3 3 ( a + 1 a ) . \frac{a^6+1}{a^3} = a^3+\frac{1}{a^3} = (a+\frac{1}{a})^3- 3(a+\frac{1}{a}).

Now, since a a is a root of the given equation, we get = a 2 + 1 = 3 a a^2+1=3a or a + 1 a = 3 a+\frac{1}{a}= 3 Plugging this value, we get a 6 + 1 a 3 = 18 \frac{a^6+1}{a^3} = 18 . Now all we want is it's reciprocal.

Nice and elegant.

Sanjeet Raria - 6 years, 5 months ago

very nice solution sir

Mardokay Mosazghi - 6 years, 4 months ago

Please explain that how our answer is right if we reciprocal the equation as you have done in first line .

Chirayu Bhardwaj - 5 years, 3 months ago
Christian Daang
Dec 25, 2014

My Solution.. :) . . .

If you don't see it clearly, then, here's the solution:

Since A is a root of x^2-3x+1=0,then we can say that a^2-3a+1 is also=to zero.

So,a^3/(a^6+1)=a^3/(a^2+1)(a^4-a^2+1)

Since a^2-3a+1=0→a^2+1=3a ;3a-1=a^2→9a^2-6a+1=a^4

=a^3/(3a)(9a^2-6a+1-(3a-1)+1)

=a^2/(3)(9a^2-9a+3)

=a^2/(9)(3a^2-3a+1)

=a^2/(9)(3a^2-a^2 )

=a^2/(9)(2a^2 )

=1/18

∴ α+β=1+18=19 ∎

NaiTik ChOksi
Dec 26, 2014

Let a,b be the 2 solns. Then a+b=3 and ab=1 ((a+b)^3)= a^3+b^3+3ab(a+b)... Substituting the values of a+b and ab, 18=a^3+b^3. 18=a^3+1/a^3 therefore a^3/1+a^6=1/18 therefore 1+18=19

Mehul Chaturvedi
Dec 24, 2014

So a 2 + 1 = 3 a a^2 + 1 = 3a

and this gives:

a a 2 + 1 = 1 3 , \dfrac{a}{a^2 + 1} = \dfrac{1}{3},

and

( a 2 + 1 ) 2 = 9 a 2 a 4 + 1 = 7 a 2 . (a^2 + 1)^2 = 9a^2 \implies a^4 + 1 = 7a^2.

So

a 2 a 4 a 2 + 1 = a 2 6 a 2 = 1 6 . \dfrac{a^2}{a^4 - a^2 + 1} = \dfrac{a^2}{6a^2} = \dfrac{1}{6}. And finally

a 3 a 6 + 1 = a a 2 + 1 a 2 a 4 a 2 + 1 = 1 3 1 6 = 1 18 . \dfrac{a^3}{a^6 + 1} = \dfrac{a}{a^2 +1}\cdot \dfrac{a^2}{a^4 - a^2 + 1} = \dfrac{1}{3}\cdot \dfrac{1}{6} = \dfrac{1}{18}.

@Mehul Chaturvedi Why don't you make a set of these good problems - Are you smater than me thanks

U Z - 6 years, 5 months ago

Log in to reply

Ok I will make it now only

Mehul Chaturvedi - 6 years, 5 months ago
Lawrence Pauls
Oct 26, 2016
  • Using the quadratic formula the roots are 3 ± 5 2 \frac{3±√5}{2}
  • pick one of the roots a = 3 + 5 2 \frac{3+√5}{2} to substitute into a ³ a + 1 \frac{a³}{a⁶+1}
  • a³ = 9+4√5
  • a⁶+1 = 162+72√5 = 18(9+4√5)
  • therefore α β \frac{α}{β} = 1 18 \frac{1}{18}
  • and α+β=19

A word of caution by me: Got it right in my second try! as Vikram Waradpande did, i too computed (a^3)+(1/a)^3 and got final [ a^3/a^6+1] as 1/18! But Your eyes must be good! root is a and we need to find (alpha)+(beta) don't treat alpha and a the same!

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