Question

[H.Nam]

1. a³ + b³ + c³ - 3abc

2. a¹⁰ + a⁵ + 1

3. a⁸ + a + 1

4. a⁸ + a⁷ + 1

5. a¹⁶ + a⁸b⁸ + b¹⁶

6. (a + 1)(a + 3)(a + 5)(a + 7) + 15

7. 4x²y² (2x + y) + y²z² ( z - y) + x²z² ( 2x + z)

8. be ( a + b)(b - c) - ac(b + d)(a - c) + ab(c + d(a - b)

9. (x - y)³ + (y - z)³ + (z - x)³

10. x⁴ + 6x³ + 7x² - 6x + 1

Answer 1

[Nguyễn Hoàng Minh]

\(1,=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\\ =\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\\ 2,=a^{10}-a+a^5-a^2+a^2+a+1\\ =a\left(a^3-1\right)\left(a^3+1\right)+a^2\left(a^3-1\right)+\left(a^2+a+1\right)\\ =\left(a-1\right)\left(a^2+a+1\right)\left(a^4+a^2+a\right)+\left(a^2+a+1\right)\\ =\left(a^2+a+1\right)\left[\left(a-1\right)\left(a^4+a^2+a\right)+1\right]\\ =\left(a^2+a+1\right)\left(a^5-a^4+a^3-a+1\right)\)

\(3,=a^8+a^7-a^7+a^6-a^6+a^5-a^5+a^4-a^4+a^3-a^3+a^2-a^2+a+1\\ =a^6\left(a^2+a+1\right)-a^5\left(a^2+a+1\right)+a^3\left(a^2+a+1\right)-a^2\left(a^2+a+1\right)+\left(a^2+a+1\right)\\ =\left(a^2+a+1\right)\left(a^6-a^5+a^3-a^2+1\right)\)

\(4,=a^8+a^7-a^6+a^6+1=a^6\left(a^2+a+1\right)-\left(a^3-1\right)\left(a^3+1\right)\\ =\left(a^2+a+1\right)\left[a^6-\left(a-1\right)\left(a^3+1\right)\right]\\ =\left(a^2+a+1\right)\left(a^6-a^4-a+a^3-1\right)\)

\(5,=\left(a^{16}+2a^8b^8+b^{16}\right)-a^8b^8=\left(a^4+b^4\right)^2-\left(a^4b^4\right)^2\\ =\left(a^4+b^4-a^4b^4\right)\left(a^4+b^4+a^4b^4\right)\\ 6,=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\\ =\left(a^2+8a+11\right)^2-16+15\\ =\left(a^2+8a+11\right)^2-1\\ =\left(a^2+8a+10\right)\left(a^2+8a+12\right)\)

Câu 7 mình làm riêng nhé

Answer 2

[Nguyễn Hoàng Minh]

\(7,=8x^3y^2+4x^2y^3+y^2z^3-y^3z^2+x^2z^2\left(2x+z\right)\\ =\left(8x^3y^2+y^2z^3\right)+\left(4x^2y^3-y^3z^2\right)+x^2z^2\left(2x+z\right)\\ =y^2\left(2x+z\right)\left(4x^2-2xz+z^2\right)+y^3\left(2x-z\right)\left(2x+z\right)+x^2z^2\left(2x+z\right)\\ =\left(2x+z\right)\left(4x^2y^2-2xyz+y^2z^2+2xy^3-2y^3z+x^2z^2\right)\)

Từ đây chịu thôi ;-;