46306 由四個代數式選取兩個可形成六個恒等式

\begin{align*} A=&\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-a)(b-c)}+\frac{c^4}{(c-b)(c-b)},\hskip 3cm~\\[6pt] B=&\dfrac{\dfrac{c^4-a^4}{c-a}-\dfrac{b^4-a^4}{b-a}}{c-b},\\[6pt] C=&\frac{\left|\begin{array}{ccc} ~1~&~a~&~a^4~\\ 1&b&b^4\\ 1&c&c^4\\ \end{array}\right|}{\left|\begin{array}{ccc} ~1~&~a~&~a^2~\\ 1&b&b^2\\ 1&c&c^2\\ \end{array}\right|},\\[6pt] D=&a^2+b^2+c^2+ab+bc+ca. \end{align*}

1. $f(x)=a^4\cdot \dfrac{(x-b)(x-c)}{(a-b)(a-c)}+b^4\cdot \dfrac{(x-a)(x-c)}{(b-a)(b-c)}+c^4\cdot \dfrac{(x-a)(x-b)}{(c-a)(c-b)}$.

2. 領導係數 $A=\dfrac{a^4}{(a-b)(a-c)}+\dfrac{b^4}{(b-a)(b-c)}+\dfrac{c^4}{(c-a)(c-b)}$.

1. $f(x)=\dfrac{\dfrac{c^4-a^4}{c-a}-\dfrac{b^4-a^4}{b-a}}{c-b}\cdot (x-b)(x-a)+\dfrac{b^4-a^4}{b-a}\cdot (x-a)+a^4$.

2. 領導係數 $B=\dfrac{\dfrac{c^4-a^4}{c-a}-\dfrac{b^4-a^4}{b-a}}{c-b}$.

1. $\left\{\begin{array}{lcl} r+q\cdot a+p\cdot a^2=a^4\\ r+q\cdot b+p\cdot b^2=b^4\\ r+q\cdot c+p\cdot c^2=c^4 \end{array}\right.$.

2. 領導係數 $p=C=\dfrac{\left|\begin{array}{ccc} ~1~&~a~&~a^4~\\ 1&b&b^4\\ 1&c&c^4\\ \end{array}\right|}{\left|\begin{array}{ccc} ~1~&~a~&~a^2~\\ 1&b&b^2\\ 1&c&c^2\\ \end{array}\right|}$.

1. $A=B$ 即 $\dfrac{a^4}{(a\!-\!b)(a\!-\!c)}+\dfrac{b^4}{(b\!-\!a)(b\!-\!c)}+\dfrac{c^4}{(c\!-\!a)(c\!-\!b)}=\dfrac{\dfrac{c^4\!-\!a^4}{c\!-\!a}\!-\!\dfrac{b^4\!-\!a^4}{b\!-\!a}}{c\!-\!b}$.

2. $A=C$ 即 $\dfrac{a^4}{(a-b)(a-c)}+\dfrac{b^4}{(b-a)(b-c)}+\dfrac{c^4}{(c-a)(c-b)}= \dfrac{\left|\begin{array}{ccc} ~1~&~a~&~a^4~\\ 1&b&b^4\\ 1&c&c^4\\ \end{array}\right|}{\left|\begin{array}{ccc} ~1~&~a~&~a^2~\\ 1&b&b^2\\ 1&c&c^2\\ \end{array}\right|}$.

3. $B=C$ 即 $\dfrac{\dfrac{c^4-a^4}{c-a}-\dfrac{b^4-a^4}{b-a}}{c-b}=\dfrac{\left|\begin{array}{ccc} ~1~&~a~&~a^4~\\ 1&b&b^4\\ 1&c&c^4\\ \end{array}\right|}{\left|\begin{array}{ccc} ~1~&~a~&~a^2~\\ 1&b&b^2\\ 1&c&c^2\\ \end{array}\right|}$.

\begin{align*} B=&\dfrac{\dfrac{c^4-a^4}{c-a}-\dfrac{b^4-a^4}{b-a}}{c-b}\\ =&\dfrac{\dfrac{(c-a)(c^3+c^2a+ca^2+a^3)}{c-a}-\dfrac{(b-a)(b^3+b^2a+ba^2+a^3)}{b-a}}{c-b}\\ =&\frac{(c^3+c^2a+ca^2+a^3)-(b^3+b^2a+ba^2+a^3)}{c-b}\\ =&\frac{(c^3-b^3)+(c^2a-b^2a)+(ca^2-ba^2)+(a^3-a^3)}{c-b}\\ =&\frac{(c-b)(c^2+cb+b^2)+(c-b)(ca+ab)+a^2(c-b)+0}{c-b}\\ =&(c^2+cb+b^2)+(ca+ab)+a^2\\ =&a^2+b^2+c^2+ab+bc+ca=D. \end{align*}

\begin{align*} C=&\dfrac{\left|\begin{array}{ccc} ~1~&~a~&~a^4~\\ 1&b&b^4\\ 1&c&c^4\\ \end{array}\right|}{\left|\begin{array}{ccc} ~1~&~a~&~a^2~\\ 1&b&b^2\\ 1&c&c^2\\ \end{array}\right|}=\dfrac{\left|\begin{array}{ccc} ~1~&~a~&~a^4~\\ 0&~b-a~&~b^4-a^4~\\ 0&c-a&c^4-a^4 \end{array}\right|}{\left|\begin{array}{ccc} ~1~&~a~&~a^2~\\ 0&~b-a~&~b^2-a^2~\\ 0&c-a&c^2-a^2 \end{array}\right|}\\ =&\dfrac{\left|\begin{array}{cc} ~b-a~&~b^4-a^4~\\ c-a&c^4-a^4 \end{array}\right|}{\left|\begin{array}{cc} ~b-a~&~b^2-a^2~\\ c-a&c^2-a^2 \end{array}\right|}=\dfrac{\left|\begin{array}{cc} ~b-a~&~(b-a)(b^3+b^2a+ba^2+a^3)~\\ c-a&(c-a)(c^3+c^2a+ca^2+a^3) \end{array}\right|}{\left|\begin{array}{cc} ~b-a~&~(b-a)(b+a)~\\ c-a&(c-a)(c+a) \end{array}\right|}\\ =&\dfrac{(b-a)(c-a)\left|\begin{array}{cc} ~1~&~b^3+b^2a+ba^2+a^3~\\ 1&c^3+c^2a+ca^2+a^3 \end{array}\right|}{(b-a)(c-a)\left|\begin{array}{cc} ~1~&~b+a~\\ 1&c+a \end{array}\right|}=\dfrac{\left|\begin{array}{cc} ~1~&~b^3+b^2a+ba^2+a^3~\\ 1&c^3+c^2a+ca^2+a^3 \end{array}\right|}{\left|\begin{array}{cc} ~1~&~b+a~\\ 1&c+a \end{array}\right|}\\ =&\dfrac{(c^3+c^2a+ca^2+a^3)-(b^3+b^2a+ba^2+a^3)}{(c+a)-(b+a)}\\ =&\frac{(c^3-b^3)+(c^2a-b^2a)+(ca^2-ba^2)+(a^3-a^3)}{c-b}\\ =&\frac{(c-b)(c^2+cb+b^2)+(c-b)(ca+ab)+a^2(c-b)+0}{c-b}\\ =&\frac{(c-b)[(c^2+cb+b^2)+(ca+ab)+a^2]}{c-b}\\ =&(c^2+cb+b^2)+(ca+ab)+a^2\\ =&a^2+b^2+c^2+ab+bc+ca=D. \end{align*}

1. 直接證:

\begin{align*} A=&\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-a)(b-c)}+\frac{c^4}{(c-b)(c-b)}\\ =&\frac{a^4(c-b)+b^4(a-c)+c^4(b-a)}{(a-b)(b-c)(c-a)}\\ =&\frac{(a^4c-ac^4)+(ab^4-a^4b)+(bc^4-b^4c)}{(a-b)(b-c)(c-a)}\\ =&\frac{ac(a^3-c^3)+ab(b^3-a^3)+bc(c^3-b^3)}{(a-b)(b-c)(c-a)}\\ =&\frac{ac(a\!-\!c)(a^2\!+\!ac\!+\!c^2)\!+\!ab(b\!-\!a)(b^2\!+\!ab\!+\!a^2)\!+\!bc(c\!-\!b)(c^2\!+\!bc\!+\!b^2)}{(a-b)(b-c)(c-a)}\\ =&\frac{ac(a-c)[(a^2+b^2+c^2+ab+bc+ca)-(b^2+ab+bc)]}{(a-b)(b-c)(c-a)}\\ &+\frac{ab(b-a)[(a^2+b^2+c^2+ab+bc+ca)-(c^2+bc+ca)]}{(a-b)(b-c)(c-a)}\\ &+\frac{bc(c-b)[(a^2+b^2+c^2+ab+bc+ca)-(a^2+ab+ca)]}{(a-b)(b-c)(c-a)}\\ =&\frac{(a^2+b^2+c^2+ab+bc+ca)[ac(a-c)+ab(b-a)+bc(c-b)]}{(a-b)(b-c)(c-a)}\\ &-\frac{ac[(a-c)(b^2+ab+bc)]}{(a-b)(b-c)(c-a)}-\frac{ab[(b-a)(c^2+bc+ca)]}{(a-b)(b-c)(c-a)}\\ &-\frac{bc[(c-b)(a^2+ab+ca)]}{(a-b)(b-c)(c-a)}\\ =&\frac{(a^2+b^2+c^2+ab+bc+ca)[ac(a-c)+b^2(a-c)+b(c^2-a^2)]}{(a-b)(b-c)(c-a)}\\ &-\frac{ac(ab^2+a^2b-b^2c-bc^2)}{(a-b)(b-c)(c-a)}-\frac{ab(bc^2+b^2c-ac^2-a^2c)}{(a-b)(b-c)(c-a)}\\ &-\frac{bc(a^2c+ac^2-a^2b-ab^2)}{(a-b)(b-c)(c-a)}\\ =&\frac{(a^2+b^2+c^2+ab+bc+ca)[ac(a-c)+b^2(a-c)-(bc+ba)(a-c)]}{(a-b)(b-c)(c-a)}\\ &-\frac{[ac(ab^2)-bc(a^2b)]+[ac(a^2b)-ab(a^2c)]+[ab(bc^2)-ac(b^2c)]}{(a-b)(b-c)(c-a)}\\ &-\frac{[ab(b^2c)-bc(ab^2)]+[bc(a^2c)-ab(ac^2)]+[bc(ac^2)-ac(bc^2)]}{(a-b)(b-c)(c-a)}\\ =&\frac{(a^2+b^2+c^2+ab+bc+ca)[(a-c)(ac+b^2-bc-ba)]}{(a-b)(b-c)(c-a)}\\ &-\frac{0+0+0}{(a-b)(b-c)(c-a)}-\frac{0+0+0}{(a-b)(b-c)(c-a)}\\ =&\frac{(a^2+b^2+c^2+ab+bc+ca)[(a-c)(b-c)(b-a)]}{(a-b)(b-c)(c-a)}\\ =&\frac{(a^2+b^2+c^2+ab+bc+ca)[(c-a)(b-c)(a-b)]}{(a-b)(b-c)(c-a)}\\ =&a^2+b^2+c^2+ab+bc+ca=D\quad \hbox{得證}. \end{align*}

2. 間接證 : 由 $A=B$ 與 $B=D$ 推得 $A=D$ 即

$$\frac{a^4}{(a\!-\!b)(a\!-\!c)}+\frac{b^4}{(b\!-\!a)(b\!-\!c)}+\frac{c^4}{(c\!-\!b)(c\!-\!b)}\!=\!a^2+b^2+c^2+ab+bc+ca.$$

3. 間接證 : 由 $A=C$ 與 $C=D$ 推得 $A=D$ 即

$$\frac{a^4}{(a\!-\!b)(a\!-\!c)}+\frac{b^4}{(b\!-\!a)(b\!-\!c)}+\frac{c^4}{(c\!-\!b)(c\!-\!b)}\!=\!a^2+b^2+c^2+ab+bc+ca.$$

1. 利用拉格朗日插值法、 牛頓插值法、 克拉瑪公式求得二次多項式的領導係數且領導係數是唯一的, 因此 $A=B=C$ 推得 $A=B$, $A=C$, $B=C$。

2.我們證得 $B=D$ 與 $C=D$。

3. 我們直接或間接證得 $A=D$。