Question

Given that \(\Large a+b=3\) and \(\Large a^2+b^2=7\), find the value of
\(\Large a^{10}+2a^9b+3a^8b^2+4a^7b^3+5a^6b^4\\\Large +6a^5b^5+5a^4b^6+4a^3b^7+3a^2b^8+2ab^9+b^{10}\)

Answer 1

Very interesting question..

Answer 2

First of all using a+b and a^2+b^2 you can find out ab

Answer 3

(a+b)^2 = (a^2+b^2) + 2ab

Answer 4

Then :

(a-b)^2 = a^2 + b^2 - 2ab

Using a^2+b^2 and ab, find out (a-b)

Answer 5

Then using a-b and a+b values we can find out a and b

Answer 6

Plug in the values and solve...

Answer 7

Well this was a primary school answer.. lets search for an olympiad type answer

Answer 8

ab=1

Answer 9

Great

Answer 10

hehe, I started thinking of binomial theorem.

Answer 11

:D @AkashdeepDeb I too but when I saw the coefficients I knew it would not be the best choice for now

Answer 12

Well we will search for a better answer later on first lets find out the answer.

Answer 13

@JungHyunRan , you have calculated ab. Now try finding out (a-b)

Answer 14

Use \[(a-b)^2 = a^2 + b^2 - 2ab\]

Answer 15

\(\Large a-b=\sqrt{5}\) right?

Answer 16

\[(a-b) = \sqrt{a^2 + b^2 - 2ab}\]

Yes.

Answer 17

I would have preferred the use of \(\pm\), in such a good question

Answer 18

ah, yup :) \(\Large a-b=\pm\sqrt{5}\)

Answer 19

\[(a-b) = \pm \sqrt{5}\]

Answer 20

Then there'll be multiple values for a and b and thus creating more values for the final answer right?

Answer 21

No then we will use the fact that ab is positive and thus a and b must have the same sign

Answer 22

Generally in such cases, the answers are a and b interchanged

Answer 23

The same would be here :)

Answer 24

Which when you notice the expression would find out that it will not affect whether a and b are interchanged :)

Answer 25

Excellent! Thanks @vishweshshrimali5 !

Answer 26

So you can take \(\pm\) or just forget about that :)

My pleasure @AkashdeepDeb :)

Answer 27

Well lets find out a and b @JungHyunRan .

We have a+b and a-b

Answer 28

Great and b ?

Answer 29

\(\Large \begin{cases} a=\frac{3+\sqrt{5}}{2} \\ b=\frac{3-\sqrt{5}}{2}\end{cases}\)

OR \(\Large \begin{cases}a=\frac{3-\sqrt{5}}{2}\\ b=\frac{3+\sqrt{5}}{2}\end{cases}\)

Answer 30

Great now select any one set because the answer would remain same whichever set you select

Answer 31

yes

Answer 32

So which one do you select ? 1st one or 2nd ?

Answer 33

I select 1st

Answer 34

Very good

Answer 35

Now lets see what the expression was

Answer 36

\[\Large a^{10}+2a^9b+3a^8b^2+4a^7b^3+5a^6b^4\\\Large
+6a^5b^5+5a^4b^6+4a^3b^7+3a^2b^8+2ab^9+b^{10}\]

Now first of all lets simplify it... use the fact that ab = 1

Answer 37

Thus, \(\large{2a^9b = 2a^8 * ab = 2a^8}\) and so on

Answer 38

So your expression would become ?

Answer 39

\(\Large (a^{10}+b^{10})+ab(2a^8+3a^7b+4a^6b^2+5a^5b^3+6a^4b^4\\\Large +5a^3b^5+4a^2b^6+3ab^7+2b^8)\)

Answer 40

Good then use the fact that ab = 1...

Answer 41

Your expression would become:

\[\Large a^{10}+2a^8+3a^6+4a^4+5a^2\\\Large +6+5b^2+4b^4+3b^6+2b^8+b^{10}\]

Answer 42

Looks much better. Doesn't it ?

Answer 43

yeah :)

Answer 44

\[\large{=(a^{10} + b^{10}) + 2(a^8 + b^8) + 3(a^6 + b^6) + 3(a^4 + b^4) + 5(a^2+b^2) + 6}\]

Answer 45

We also know (a^2+b^2) value so lets plug in that too.

What will we get ?

Answer 46

So, how to find value of \(\Large a^{10}+b^{10}\) ?

Answer 47

Lets start in increasing order.

Answer 48

First lets evaluate \(a^2 + b^2\) but we already know that it is 7

Answer 49

Then lets find out a^4 + b^4

Answer 50

\(\Large a^4+b^4=(a^2+b^2)^2-2a^2b^2=7^2-2=47\)

Answer 51

Great !!

Answer 52

Then \(\large{a^6+b^6}\) ?

Answer 53

Try using (a^4+b^4) and (a^2+b^2)

Answer 54

Or better use (a^3+b^3)^2

Answer 55

First find out (a^3+b^3):

\[\large{a^3+b^3 = (a+b)(a^2+b^2-ab) = 3(7-1) = 18}\]

Answer 56

\(\Large a^6+b^6=(a^3+b^3)^2-2a^3b^3=18^2-2=322\)

Answer 57

Very good

Answer 58

Now we have :

(a^8 + b^8)

Answer 59

Use (a^4+b^4)

Answer 60

\(\Large =(a^4+b^4)^2-2a^4b^4=47^2-2=2207\)

Answer 61

Great

Answer 62

Now only (a^10 + b^10)

Answer 63

\[(a^5 + b^5) = (a^2 + b^2)(a^3+b^3) - a^2b^3 - b^2a^3 = (a^2+b^2)(a^3+b^3) - a^2 b^2(a+b)\]

Answer 64

Now using this find out (a^5+b^5)

Answer 65

\(\Large a^5+b^5=7\times 18-3=123\)

Answer 66

Great

Answer 67

Thus, \(\Large a^{10}+b^{10}=(a^5+b^5)^2-2a^5b^5=123^2-2=15127\)

Answer 68

Fantastic

Answer 69

Now lets simplify final answer

Answer 70

What will we get ?

Answer 71

20736

Answer 72

Great (I have not checked it using calculator so I am trusting you on it) :)

Answer 73

Fantastic work @JungHyunRan . I enjoyed helping you :)

Answer 74

Thank you :D

Answer 75

Your welcome :)

Good day