\(x^3+3xy+y^3-1\)

\(=x^3+3xy\left(x+y\right)+y^3-3xy\left(x+y\right)+3xy-1\)

\(=\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left[\left(x+y\right)^2+\left(x+y\right)+1\right]-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left(x^2+y^2-xy+x+y+1\right)\)

\(x^3+3xy+y^3-1\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy-1\)

\(=\left(x+y-1\right)\left[\left(x+y\right)^2+\left(x+y\right)+1\right]-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left[x^2+2xy+y^2+x+y-3xy+1\right]\)

\(=\left(x+y-1\right)\left(x^2+y^2+x+y-xy+1\right)\)

\(a^6+a^4+a^2b^2+b^4-b^6\)

\(=a^6-b^6+a^4+2a^2b^2+b^4-a^2b^2\)

\(=\left(a-b\right)\left(a+b\right)\left(a^2+ab+b^2\right)\left(a^2-ab+b^2\right)+\left(a^2+b^2\right)^2-\left(ab\right)^2\)

\(=\left(a^2+b^2+ab\right)\left(a^2+b^2-ab\right)\left(a^2-b^2+1\right)\)

a⁶ + a⁴ + a²b² + b⁴ - b⁶

= (a⁶ - b⁶) + (a⁴ + a²b² + b⁴)

= [(a²)³ - (b²)³] + (a⁴ + a²b² + b⁴)

= (a² - b²)(a⁴ + a²b² + b⁴) + (a⁴ + a²b² + b⁴)

= (a - b)(a + b)(a⁴ + a²b² + b⁴) + (a⁴ + a²b² + b⁴)

= (a⁴ + a²b² + b⁴)[(a - b)(a + b) + 1]

\(64x^4+y^4\)

\(=64x^4+16x^2y^2+y^4-16x^2y^2\)

\(=\left(8x^2\right)^2+2.8x^2.y^2+y^4-\left(4xy\right)^2\)

\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)

\(=\left(8x^2-4xy-y^2\right)\left(8x^2+4xy-y^2\right)\)

64x⁴ + 16x²y² + y⁴ - 16x²y²

= (8x²)² + 2.8x².y² + (y²)² - (4xy)²

= (8x² + y²)² - (4xy)²

= (8x² - 4xy + y²)(8x² + 4xy + y²)

Từ điều kiện đã cho không tính được $x+y$ bạn nhé. Bạn xem lại đề.

\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)

\(=3\left(x^4+x^2+1\right)-\left(x^4+x^2+1+2x^3+2x^2+2x\right)\)

\(=2\left(x^4+x^2+1\right)-2\left(x^3+x^2+x\right)\)

\(=2\left(x^4+x^2+1-x^3-x^2-x\right)\)

\(=2\left(x^4-x^3-x+1\right)\)

\(=2\left(x^3\left(x-1\right)-\left(x-1\right)\right)\)

\(=2\left(x-1\right)\left(x^3-1\right)\)

\(=2\left(x-1\right)^2\left(x^2+x+1\right)\)

\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)

\(=3\left[x^4+2x^2+1-x^2\right]-\left(x^2+x+1\right)^2\)

\(=3\left[\left(x^2+1\right)^2-x^2\right]-\left(x^2+x+1\right)^2\)

\(=3\left(x^2+x+1\right)\left(x^2-x+1\right)-\left(x^2+x+1\right)^2\)

\(=\left(x^2+x+1\right)\left(3x^2-3x+3-x^2-x-1\right)\)

\(=\left(x^2+x+1\right)\left(2x^2-4x+2\right)\)

\(=2\left(x-1\right)^2\cdot\left(x^2+x+1\right)\)

\(4x^4-32x^2+1\)

\(=4x^4+12x^3+2x^2-12x^3-36x^2-6x+2x^2+6x+1\)

\(=2x^2\left(2x^2+6x+1\right)-6x\left(2x^2+6x+1\right)+\left(2x^2+6x+1\right)\)

\(=\left(2x^2-6x+1\right)\left(2x^2+6x+1\right)\)

4x⁴ - 32x² + 1

= (2x²)² - 2.2x².8 + 64 - 63

= (2x² - 8)² - (3√7)²

= (2x² - 8 - 3√7)(2x² - 8 + 3√7)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)

\(=\left(x^2+7x+11\right)^2-25\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(y=x^2+7x+10\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(x^2+7x+10+1-5)(x^2+7x+10+1+5)\\=(x^2+7x+6)(x^2+7x+16)\\=(x^2+x+6x+6)(x^2+7x+16)\\=[x(x+1)+6(x+1)](x^2+7x+16)\\=(x+1)(x+6)(x^2+7x+16)\\Toru\)

\(x^2+2xy+y^2-x-y-12\)

\(=\left(x+y\right)^2-\left(x+y\right)-12\)

\(=\left(x+y\right)^2-4\left(x+y\right)+3\left(x+y\right)-12\)

\(=\left(x+y\right)\left(x+y-4\right)+3\left(x+y-4\right)\)

\(=\left(x+y-4\right)\left(x+y+3\right)\)

Áp dụng định lý Pytago

\(AC^2=BC^2-AB^2\)

\(=10^2-6^2=64\)

\(\Rightarrow AC=8cm\)

\(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)

\(=\left(9x^2-6x+4\right)\left(3x-1\right)\)

\(27x^3-27x^2+18x-4\\=27x^3-9x^2-18x^2+6x+12x-4\\=9x^2(3x-1)-6x(3x-1)+4(3x-1)\\=(3x-1)(9x^2-6x+4)\)