問題
次の式を計算せよ.
解説
- (1) $(x – 1)(x – 3)(x + 1)(x + 3)$
- $=(x-1)(x+1)(x-3)(x+3)$
 $=(x^2-1)(x^2-9)$
 $=$$x^4-10x^2+9$
 
- $=(x-1)(x+1)(x-3)(x+3)$
- (2) $(x + 2)(x + 5)(x – 4)(x – 1)$
- $=(x+2)(x-1)(x+5)(x-4)$
 $=(x^2+x-2)(x^2+x-20)$
 $=(x^2+x)^2-22(x^2+x)+40$
 $=$$x^4+2x^3-21x^2-22x+40$
 
- $=(x+2)(x-1)(x+5)(x-4)$
- (3) $(a – b)(a + b)(a^2 + b^2)(a^4 + b^4)$
- $=(a^2-b^2)(a^2+b^2)(a^4+b^4)$
 $=(a^4-b^4)(a^4+b^4)$
 $=$$a^8-b^8$
 
- $=(a^2-b^2)(a^2+b^2)(a^4+b^4)$
- (4) $(2x – y)^3(2x + y)^3$
- $=\{(2x-y)(2x+y)\}^3$
 $=(4x^2-9y^2)^3$
 $=$$64x^6-48x^4y^2+12x^2y^4-y^6$
 
- $=\{(2x-y)(2x+y)\}^3$
- (5) $(a + b)^2(a – b)^2(a^4 + a^2b^2 + b^4)^2$
- $=\{(a-b)(a+b)(a^4+a^2b^2+b^4)\}^2$
 $=\{(a^2-b^2)(a^4+a^2b^2+b^4)\}^2$
 $=(a^6-b^6)^2$
 $=$$a^{12}-2a^6b^6+b^{12}$
 
- $=\{(a-b)(a+b)(a^4+a^2b^2+b^4)\}^2$
- (6) $(x + 2)(x – 2)(x^2 + 2x + 4)(x^2 – 2x + 4)$
- $=(x+2)(x^2 – 2x + 4)(x – 2)(x^2 + 2x + 4)$
 $=(x^3+8)(x^3-8)$
 $=$$x^6-64$
 
- $=(x+2)(x^2 – 2x + 4)(x – 2)(x^2 + 2x + 4)$
- (7) $(a + b + c)^2 + (a + b – c)^2 + (b + c – a)^2 + (c + a – b)^2$
- $b+c=m, b-c=n$と置くと,
 $(a+m)^2+(a+n)^2+(m-a)^2+(a-n)^2$
 $=a^2+2am+m^2+a^2+2an+n^2+m^2-2am+a^2+a^2-2an+n^2$
 $=4a^2+2(m^2+n^2)$
- ここで$m, n$を戻すと,
 $4a^2+2\{(b+c)^2+(b-c)^2\}$
 $=4a^2+2(b^2+2bc+c^2+b^2-2bc+c^2)$
 $=$$4a^2+4b^2+4c^2$
 
- $b+c=m, b-c=n$と置くと,
 
  
  
  
  