我的作業...有關於因式分解的數學

x³ + 6x²y + 12xy² + 8y³
= x³ + 3x²(2y) + 3x(2y)² + (2y)³
= (x + 2y)³


8x³ - 36x²y + 54xy² - 27y³
= (2x)³ - 3(2x)²(3y) + 3(2x)(3y)² - (3y)³
= (2x - 3y)³


令 u = x² + 2x - 6
則 (x² + 2x + 4)(x² + 2x - 6) + 21
= (u + 10)u + 21
= u² + 10u + 21
= (u + 3)(u + 7)
= (x² + 2x - 6 + 3)( x² + 2x - 6 + 7)
= (x² + 2x - 3)( x² + 2x + 1)
= (x - 1)(x + 3)(x + 1)²


令 u = x² + 8x + 7
則 (x + 1)(x + 3)(x + 5)(x + 7) + 15
= [(x + 1)(x + 7)] [(x + 3)(x + 5)] + 15
= (x² + 8x + 7)(x² + 8x + 15) + 15
= u(u + 8) + 15
= u² + 8u + 15
= (u + 3)(u + 5)
= (x² + 8x + 7 + 3)( x² + 8x + 7 + 5)
= (x² + 8x + 10)(x² + 8x + 12)


令 u = x² + 5x + 6
則 (x + 1)(x + 2)(x + 3)(x + 6) - 15x²
= [(x + 1)(x + 6)] [(x + 2)(x + 3)] - 15x²
= (x² + 7x + 6)(x² + 5x + 6) - 15x²
= (u + 2x)u - 15x²
= u² + 2xu - 15x²
= (u - 3x)(u + 5x)
= (x² + 5x + 6 - 3x)( x² + 5x + 6 + 5x)
= (x² + 2x + 6)( x² + 10x + 6)


(xy + 1)(x + 1)(y + 1) + xy
= (xy + 1)(xy + x + y + 1) + xy
= (xy + 1) [(xy + x + 1) + y] + xy
= (xy + 1)(xy + x + 1) + (xy + 1)y + xy
= (xy + 1)(xy + x + 1) + y[(xy + 1) + x]
= (xy + 1)(xy + x + 1) + y(xy + x + 1)
= (xy + x + 1) [(xy + 1) + y]
= (xy + x + 1)(xy + y + 1)


令 u = x² + 2x - 2
則 (x² + 2x - 2)( x² + 2x + 7) + 18
= u(u + 9) + 18
= u² + 9u + 18
= (u + 3)(u + 6)
= (x² + 2x - 2 + 3)( x² + 2x - 2 + 6)
= (x² + 2x + 1)( x² + 2x + 4)
= (x + 1)²(x² + 2x + 4)