x to the power of 5 added to reciprocal of x to the power of 5, when x added to the reciprocal of x is 5
Solution
Given:
x+x1=5
We need to find:
x5+x51
Let's denote y=x+x1. Then, y=5.
To find x2+x21, we square y:
(x+x1)2=x2+x21+2y2=x2+x21+252=x2+x21+225=x2+x21+2x2+x21=23
Next, we find x3+x31 by multiplying x+x1 and x2+x21:
(x+x1)(x2+x21)=x3+x31+x+x1y⋅(x2+x21)=x3+x31+y5⋅23=x3+x31+5115=x3+x31+5x3+x31=110
Next, we find x4+x41 by squaring x2+x21:
(x2+x21)2=x4+x41+2232=x4+x41+2529=x4+x41+2x4+x41=527
Finally, we find x5+x51 by multiplying x+x1 and x4+x41:
(x+x1)(x4+x41)=x5+x51+x3+x31y⋅(x4+x41)=x5+x51+x3+x315⋅527=x5+x51+1102635=x5+x51+110x5+x51=2525
So, x5+x51=2525.
Would you like any details or have any questions?
Here are 5 questions you might ask next:
How do you solve other similar polynomial equations?
Can you explain the method used to solve for x3+x31 in more detail?
How can these methods be applied to other types of equations?
What are some common applications of these types of mathematical problems?
Can you show the derivation of the binomial theorem used in such expansions?
Tip: When dealing with expressions involving reciprocals and powers, leveraging identities and symmetry in algebra can simplify the problem significantly.