簡約 15(a-b+c)^2+(b-c+a)^2+2(a-b+c)(b-c+a)

解

簡約

15 (a - b + c)^{2} + (b - c + a)^{2} + 2 (a - b + c) (b - c + a)

15 (a - b + c)^{2} - b^{2} + 3 a^{2} + 2 b c + 2 ab - c^{2} - 2 a c

解答ステップ

15 (a - b + c)^{2} + (b - c + a)^{2} + 2 (a - b + c) (b - c + a)

拡張

(b - c + a)^{2} : b^{2} - 2 b c + 2 ab + c^{2} + a^{2} - 2 a c

= 15 (a - b + c)^{2} + b^{2} - 2 b c + 2 ab + c^{2} + a^{2} - 2 a c + 2 (a - b + c) (b - c + a)

拡張

2 (a - b + c) (b - c + a) : 2 a^{2} - 2 b^{2} + 4 b c - 2 c^{2}

= 15 (a - b + c)^{2} + b^{2} - 2 b c + 2 ab + c^{2} + a^{2} - 2 a c + 2 a^{2} - 2 b^{2} + 4 b c - 2 c^{2}

条件のようなグループ

= 15 (a - b + c)^{2} + b^{2} - 2 b^{2} + 2 a^{2} + a^{2} - 2 b c + 4 b c + 2 ab + c^{2} - 2 c^{2} - 2 a c

類似した元を足す:

b^{2} - 2 b^{2} = - b^{2}

= 15 (a - b + c)^{2} - b^{2} + 2 a^{2} + a^{2} - 2 b c + 4 b c + 2 ab + c^{2} - 2 c^{2} - 2 a c

類似した元を足す:

2 a^{2} + a^{2} = 3 a^{2}

= 15 (a - b + c)^{2} - b^{2} + 3 a^{2} - 2 b c + 4 b c + 2 ab + c^{2} - 2 c^{2} - 2 a c

類似した元を足す:

- 2 b c + 4 b c = 2 b c

= 15 (a - b + c)^{2} - b^{2} + 3 a^{2} + 2 b c + 2 ab + c^{2} - 2 c^{2} - 2 a c

類似した元を足す:

c^{2} - 2 c^{2} = - c^{2}

= 15 (a - b + c)^{2} - b^{2} + 3 a^{2} + 2 b c + 2 ab - c^{2} - 2 a c