if x+(1/x)= 3 find x^5+(1/x^5)
Please explain it with steps and NO SPAM ANSWERS otherwise I'll report you.
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(x+1/x)^5 = x^5 + 5 x^3 + 10 x + 10/x + 5 /x^3 + 1/ x^5
x^5 + 1/x^5 = (x+1/x)^5 - 5 (x^3 + 1/x^3) - 10 (x+1/x)
(x+1/x)^3 = x^3 + 1/x^3 + 3 (x+1/x)
x^3 + 1/x^3 = (x+1/x)^3 - 3 (x+1/x) = 27 - 9 = 18
So, x^5 + 1/x^5 = 3^5 - 5 (18) - 10 (3) = 243 - 90 - 30 = 123
x^5 + 1/x^5 = (x+1/x)^5 - 5 (x^3 + 1/x^3) - 10 (x+1/x)
(x+1/x)^3 = x^3 + 1/x^3 + 3 (x+1/x)
x^3 + 1/x^3 = (x+1/x)^3 - 3 (x+1/x) = 27 - 9 = 18
So, x^5 + 1/x^5 = 3^5 - 5 (18) - 10 (3) = 243 - 90 - 30 = 123
Anonymous:
you know binomial theorem. right??
No I'm in class 9. Would you please tell me that?
how did you expand that big one then?
did you multiply everytime
Are you talking about identities like (a+b)^2= a^2+b^2+2ab and (a-b) ^2= a^2+b^2-2ab etc.
did you expand (x +1/x)^5??
No I didn't. I can only go for something like (x+1/x)^3. (x+1/x)^5 seems a bit tough to expand. If you know how expand it, please tell me.
just go to youtube and learn binomial theorem.. it will help a lot.. i used the same her
same here*
Oh thanks for your suggestion. I'll surely go for it.
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