52번 어케 하는 거야...? 고1 수학인데 어려워서

 

모르겠어

[익인1]

헉 쓰니 이미 풀었을지 모르겠지만...!!

1)
8x³-27y³-18xy-1

= (2x)³+(-3y)³+(-1)³-18xy

= (2x-3y-1){(2x)²+(-3y)²+(-1)²-(2x)(-3y)-(-3y)(-1)-(-1)(2x)}

= (2x-3y-1)(4x²+9y²+1+6xy-3y+2x)

[익인2]

2)
(x-1)³+(x-2)³-(2x-3)³

= {(x-1)³+(x-2)³}-(2x-3)³

= {(x-1)+(x-2)}{(x-1)²-(x-1)(x-2)+(x-2)²}-(2x-3)³

= (2x-3)(x²-3x+3)-(2x-3)³

= (2x-3){(x²-3x+3)-(2x-3)²}

= (2x-3)(x²-3x+3-4x²+12x-9)

= (2x-3)(-3x²+9x-6)

= -3(2x-3)(x²-3x+2)

= -3(2x-3)(x-1)(x-2)

[익인3]

3)
(a-b-1)³+(b-c-2)³+(c-a+3)³

= {(a-b-1)³+(b-c-2)³}+(c-a+3)³

= {(a-b-1)+(b-c-2)}{(a-b-1)²-(a-b-1)(b-c-2)+(b-c-2)²}+(c-a+3)³

= (a-c-3){(a-b-1)²-(a-b-1)(b-c-2)+(b-c-2)²}+(c-a+3)³

= (a-c-3)(a²+b²+1-2ab+2b-2a-ab+ac+2a+b²-bc-2b+b-c-2+b²+c²+4-2bc+4c-4b)-(a-c-3)³

= (a-c-3)(a²+3b²+c²-3ab-3bc+ac-3b+3c+3)-(a-c-3)³

= (a-c-3)(a²+3b²+c²-3ab-3bc+ac-3b+3c+3-a²-c²-9+2ac-6c+6a)

= (a-c-3)(3b²-3ab-3bc+3ac-3b-3c-6+6a)

= 3(a-c-3)(b²-ab-bc+ac-b-c-2+2a)

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

b-a +1

b-c -2

= 3(a-c-3)(b-a+1)(b-c-2)