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,