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他的回答已经明确这个多项式怎么在实数域或者复数域上分解了，我简单提一下怎么在有理数域或者整数环上分解。 由于这个多项式 x^n-y^n 是齐次的，所以我们只需要考虑 f (x) = x^n-1 这个多项式怎么分解就好了。 由著名的高斯引理，我们在 \mathbb Q [x] 还是在 \mathbb Z [x] 上进行分解是一样的。 这时候这个多项式分解成不可约多项式的结果即是著名的分圆多项式： x^n - 1 = \prod_ {d \mid n} \Phi_d (x) 关于这一点，基本上任何讲分圆多项式的时候都会提到。 比如可以在知乎 …

Write out your expanded expression in two rows multiplying first by x to get the first row and then by − y to get the second row. If you write the second row so that expressions with the same power of x are underneath each other, and do an extra term or two at the beginning and the end, you should detect quite easily what has gone wrong.

Sep 27, 2009 · E.g. x*x^(n-2)*y cancels y*x^(n-1), x*x^(n-3)*y^2 cancels y*x^(n-2)*y. I know you can't write out all of the terms. You'll have to use the '...' to express what you mean.

x^n-y^n =(x-y)[x^(n-1)+x^(n-2)y+x^(n-3)y^2+...+x^2y(n-3)+xy^(n-2)+y^(n-1)] 注意：即中括号内每项次数和为n-1，将x-y乘进去，即可直接验证。 如果是加的话， n为奇数时，可直接将 x^n+y^n=x^n-（-y）^n， n为偶数时，恒有 x^n+y^n>0 ，无实根，实数域内无法分解，如x^2+y^2

I am looking for a general expansion of $x^{n}-y^{n}$ with $x,y>0$ and $n$ being real. I came across the following formula (Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{...

Way to show $x^n + y^n = z^n$ factorises as $(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) = z^n$

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能否分解成除了第一项因式（x-y）之外，其余部分全部由（x+y）、（x*y）和它们的组合所构成的形式？即令a…

影视仓. 片头片尾快捷标记，最新exo支持软硬解，无缝换线路换源