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Hi - I know this is an old question but I thought this was a point worth making. There's actually no way to correctly evaluate a^5+1/a^5 as the two given equations are inconsistent, as you can see if you solve them individually for a (you can do this just by multiplying up and regarding them as 'disguised quadratics') - you get very different solutions. Another way to see this is by the fact that the working given below finding a+1/a=2 is correct, but if you square this equation then you get (a+1/a)^2= a^2+1/a^2+2 = 5 + 2 = 7 ~= 4 = 2^2.  

Hello nayeemtufat,

Now assuming you'll know the expressions:

a2+b2 =(a+b)2- 2ab   -------------------------- (eq. 1)

a3+b3= (a+b)3-3ab(a+b)---------------------- (eq. 2) 

Follow the steps given by Raquel and you arrive at

 a5 +(1/a5) = 40 -(a+(1/a)) ,  ------------------------------- (eq. 3)

To find a+1/a;

a2 + 1/a2 = 5

a2 + 1/a2= (a+ 1/a)2 - 2(a *1/a)

               = (a + 1/a)2  - 2

Which implies;

(a+1/a)2  -2 = 5

Isolating a + 1/a, we get 

(a + 1/a)2 = 7 ----------------------------------- (eq. 4)

Next, 

a3 + 1/a3 = 8

from the eqn 2;

a3 + 1/a3 = (a +1/a)3 - 3( a*1/a)(a + 1/a)

                = (a+1/a)3 - 3(a + 1/a)

On the RHS, we will take out (a +1/a) and then interchanging LHS and RHS we get ;

(a +1/a) ((a+1/a)2 -3) = a3 +1/a3

(a +1/a) ((a+1/a)2 -3) = 8   ------------------- ( a3 +1/a3 = 8)

(a+1/a)(7 - 3) = 8               -------------------- ( from eq.4 we know (a+ 1/a)2 = 7)

(a + 1/a) (4) = 8

(a + 1/a) = 8/4

(a+1/a) = 2  ---------------------------------- (eq .5)

put a + 1/a = 2 in eq.3

We get

a5 + 1/a5 = 40 -2

a5 + 1/a5 = 38

Multiply 1st and 2nd equation u will get

LHS= a^5+(1/a^5)+a+(1/a)

RHS = 40 now we have to find a^5 +(1/a^5)

So it will be a^5 +(1/a^5) = 40 -a+(1/a)

Now we have to find the value of a +(1/a)

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

{a+(1/a)}^2=5+2=7

Now use {a^3+(1/a^3)}= a+(1/a)}((a+(1/a))^2 -3)

a+(1/a) =8/(7-3) =8/4 =2

Now put the value in this equation

a^5 +(1/a^5) = 40 -a+(1/a) =40-2 =38



Multiply 1st and 2nd equation u will get

LHS= a^5+(1/a^5)+a+(1/a)

RHS = 40 now we have to find a^5 +(1/a^5)

So it will be a^5 +(1/a^5) = 40 -a+(1/a)

Now we have to find the value of a +(1/a)

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

{a+(1/a)}^2=5+2=7

Now use {a^3+(1/a^3)}= a+(1/a)}((a+(1/a))^2 -3)

a+(1/a) =8/(7-3) =8/4 =2

Now put the value in this equation

a^5 +(1/a^5) = 40 -a+(1/a) =40-2 =38



I thought afterwards about simplifying further. I see you've already done it, Sherin. Well spoted.

Lets solve this problem:

Now assuming you'll know the expressions:

a2+b2 =(a+b)2- 2ab   -------------------------- (eq. 1)

a3+b3= (a+b)3-3ab(a+b)---------------------- (eq. 2) 

Follow the steps given by Raquel and you arrive at

 a5 +(1/a5) = 40 -(a+(1/a)) ,  ------------------------------- (eq. 3)

To find a+1/a;

a2 + 1/a2 = 5

a2 + 1/a2= (a+ 1/a)2 - 2(a *1/a)

               = (a + 1/a)2  - 2

Which implies;

(a+1/a)2  -2 = 5

Isolating a + 1/a, we get 

(a + 1/a)2 = 7 ----------------------------------- (eq. 4)

Next, 

a3 + 1/a3 = 8

from the eqn 2;

a3 + 1/a3 = (a +1/a)3 - 3( a*1/a)(a + 1/a)

                = (a+1/a)3 - 3(a + 1/a)

On the RHS, we will take out (a +1/a) and then interchanging LHS and RHS we get ;

(a +1/a) ((a+1/a)2 -3) = a3 +1/a3

(a +1/a) ((a+1/a)2 -3) = 8   ------------------- ( a3 +1/a3 = 8)

(a+1/a)(7 - 3) = 8               -------------------- ( from eq.4 we know (a+ 1/a)2 = 7)

(a + 1/a) (4) = 8

(a + 1/a) = 8/4

(a+1/a) = 2  ---------------------------------- (eq .5)

put a + 1/a = 2 in eq.3

We get

a5 + 1/a5 = 40 -2

a5 + 1/a5 = 38



thanks a lot

1. Multiply both equations to get a new equation:

(a2+(1/a2))•(a3+(1/a3))=5•8

2. To open brackets, multiply each of the terms in the first brackets with each term of the second bracket

a5+(a2/a3)+(a3/a2)+(1/a5)=40

3. Isolate a5+(1/a5) in one side of the expression

a5+(1/a5)=40-((a3/a2)+(a2/a3))

4. Simplify fractions on the right expression

a5+(1/a5)=40-(a+(1/a))

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