Just observed this , I founded it nice so I am sharing it

Now suppose this is the question -

\(2x^{8} - 9x^{7} + 20x^{6} - 33x^{5} + 46x^{4} - 66x^{3} + 80x^{2} - 72x + 32 =0\)

Let's go in a generalized manner, if we had this equation

\( a,b,c...,y>0 , z \in R - \{0\}\)

\( ax^{2m} + bx^{2m - 1} + cx^{2m -2} + .... + zx^{m} + ...... + cx^2 + bx + a =0\)

Keeping \(x=0 , ~gives~a=0~so~we~can't~take~x=0\)

Now see the beauty,

Dividing the whole equation by \(x^m\)

\( ax^m + bx^{m-1} + cx^{m-2} + ..... + z + ....... + \dfrac{c}{x^{m-2}} + \dfrac{b}{x^{m-1}} + \dfrac{a}{x^m}\)

\( = a(x^m + \dfrac{1}{x^m}) + b(x^{m-1} + \dfrac{1}{x^{m-1}}) + c(x^{m-2} + \dfrac{1}{x^{m-2}}) + .....+ z =0\)

\( x + \dfrac{1}{x} = y\)

**Example(Application) -**

\( 2x^4 + x^3 + x^2 + x + 2 = 0\)

**See here the coefficients of \(x^n\) are a kind of mirror image after \(x^2\)**

**Dividing by \(x^2\)** ( \(x \neq 0 ~as~ 2\neq 0\))

\( 2(x^2 + \dfrac{1}{x^2}) + (x + \dfrac{1}{x}) + 1 =0\)

**Thus**

\( 2(y^2 - 2) + y + 1 =0\)

\( 2y^2 + y - 3 =0\)

\( (2y + 3)(y - 1) = 0\)

\( y = \dfrac{-3}{2} , y =1\)

**Further we have to substitute y and solve the quadratic**

**First few**

\(x^2 + \dfrac{1}{x^2} = y^2 - 2\) (**See how lovely this ratio is!** \(~ (x + \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} + \boxed{2x\times\dfrac{1}{x}}\))

\(x^3 + \dfrac{1}{x^3} = y(\boxed{y^2-2}) - \boxed{y} = y^3 - 3y\)

\(x^4 + \dfrac{1}{x^4} = y(\boxed{ y^3 - 3y}) - \boxed{y^2-2}\)

**Thus, we can generalize it as -**

\( x^{n + 1} + \dfrac{1}{x^{n+1}} = \Big(x^n + \dfrac{1}{x^n} \Big)\Big(x + \dfrac{1}{x}\Big) - \Big( x^{n - 1} + \dfrac{1}{x^{n-1}}\Big)\)

**Writing in terms of t(as most of us write)**

\( t_{n} = x^n + \dfrac{1}{x^n}\)

\(\huge{\boxed{t_{n+1} = t_{1}(t_{n}) - t_{n-1}}}\)

**The question first I asked , solve it yourself - it can be solved by the same method**

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## Comments

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TopNewestNice Note , Megh , Thanks For Sharing This ! \(\ddot\smile\)

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FYI, This is called as Reciprocal Equation. Good find! :)

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Thanks! I used this trick in a problem in Hall&Knight.

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Yay! Me too :)

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nice...

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Nice note @megh choksi . I never used this method in general. For quartics, I love this! It would help me a lot!

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See how this method is useful in solving this question

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Nice note! Did you mean \(t_{n+1} = t_{1}t_{n} - t_{n-1}\)?

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Yes , thanks for pointing out and reading it

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I just realized that it can be proven easily by \(a^{n+1} + b^{n+1} = (a+b)(a^{n}+b^{n}) - ab(a^{n-1}+b^{n-1})\).

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You're missing an x in the third equation, it should be \(cx^{2m-2}\), not just \(c^{2m-2}\) unless that was intentional.

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