這個算國中乘法公式單元的題目嗎???

Question

已知a+b+c=17
a^2+b^2+c^2=131
a^3+b^3+c^3=1073
求1/a+1/b+1/c=???
a^4+b^4+c^4=???

Answer 1

(a+b+c)^2=a^2 + b^2 + c^2 + 2ab + 2bc + 2ac
(a+b+c)^3=a^3 + b^3 +c ^3 + 6abc + 3ab^2 + 3bc^2 + 3ac^2 + 3a^2b + 3b^2c + 3a^2c
嗯～超出國中範圍
(1)
(a+b+c)^2-(a^2+b^2+c^2)=17^2-131=289-131=158
(a^2 + b^2 + c^2 + 2ab + 2bc + 2ac)-(a^2+b^2+c^2)=2ab + 2bc + 2ac=158
ab+bc+ac=158/2=79

(a+b+c)^3-(a^3+b^3+c^3)=17^3-1073=4913-1073=3840
6abc + 3ab^2 + 3bc^2 + 3ac^2 + 3a^2b + 3b^2c + 3a^2c =3840
2abc + ab^2 + bc^2 + ac^2 + a^2b + b^2c + a^2c =1280

(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+ab^2 + bc^2 + ac^2 + a^2b + b^2c + a^2c =17*131=2227
1073+(ab^2 + bc^2 + ac^2 + a^2b + b^2c + a^2c) =2227
ab^2 + bc^2 + ac^2 + a^2b + b^2c + a^2c =1154

2abc+1154=1280
2abc=126
abc=63

(ab+bc+ac)/abc=[[先除a]]{b+(b+c)/a+c}/bc=>[[再除b]]{1+(c/a)+(c/b)}/c=>[[再除c]]=>(1/c)+(1/a)+(1/b)
1/a+1/b+1/c=(ab+bc+ac)/abc=79/63

(2)
(a^3+b^3+c^3)(a+b+c)=1073*17=18241
(a^3+b^3+c^3)(a+b+c)=(a^4+b^4+c^4)+ab^3+ac^3+a^3b+bc^3+a^3c+b^3c=18241

(a^2+b^2+c^2)(ab+bc+ab)=131*79=10349
a^3b+a^2bc+a^3c+ab^3+b^3c+ab^2c+abc^2+ac^3=10349
a^3b+a^3c+ab^3+b^3c+ac^3+(abc)(a+b+c)=10349
a^3b+a^3c+ab^3+b^3c+ac^3=9278

a^4+b^4+c^4=18241-9278=8963


好複雜～(@[]@)

Answer 2

a+b+c=17
a^2+b^2+c^2+2ab+2bc+2ac=289
ab+bc+ac=79
a^3+b^3+c^3+ab^2+ac^2+a^2b+bc^2+ca^2+b^2c=2227
ab^2+ac^2+a^2b+bc^2+a^2c+b^2c=1154
(a+b+c)^3=a^3+b^3+c^3+3ab^2+3a^2b+3a^2c+3ac^2+3b^2c+3bc^2+6abc=4913
3ab^2+3a^2b+3a^2c+3ac^2+3b^2c+3bc^2+6abc=3840
6abc=378
abc=63
1/a+1/b+1/c=(ab+bc+ca)/abc=79/63
(ab+bc+ac)*(a^2+b^2+c^2)=(131*79)
a^3b+a^2bc+a^3c+ab^3+b^3c+ab^2c+abc^2+ac^3=10349
a^3b+a^3c+ab^3+b^3c+ac^3+(abc)*(a+b+c)=10349
a^3b+a^3c+ab^3+b^3c+ac^3=9278
(a^4+b^4+c^4)+ab^3+ac^3+a^3b+bc^3+a^3c+b^3c=18241
a^4+b^4+c^4=8963

基本上計算過程就是這樣....

Answer 3

a=9, b=1, c=7
1/9+1/1+1/7=1.25396825
9^4+1^4+7^4=8963