Question

Factor completely x^6 - y^6

Answer 1

`\large\boxed{x^6-y^6=(x-y)(x+y)(x^2+y^2-xy)(x^2+y^2+xy)}`

Explanation:

`x^6-y^6=x^((2)(3))-y^((2)(3))\qquad\text{use}\ (a^n)^m=a^(nm)\\\\=(x^2)^3-(y^2)^3\qquad\text{use}\ a^3-b^3=(a-b)(a^2+ab+b^2)\\\\=(x^2-y^2)\bigg((x^2)^2+x^2y^2+(y^2)^2\bigg)\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=(x-y)(x+y)\bigg((x^2)^2+2x^2y^2+(y^2)^2-x^2y^2\bigg)\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\=(x-y)(x+y)\bigg((x^2+y^2)^2-x^2y^2\bigg)\\\\=(x-y)(x+y)\bigg((x^2+y^2)^2-(xy)^2\bigg)\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=(x-y)(x+y)(x^2+y^2-xy)(x^2+y^2+xy)`

Answer 2

(x²)³ - (y²)³ = (x² - y²)(x^4 + x²y² + y^4)

Explanation:

x^6 - y^6 is the difference of two cubes: (x²)³ - (y²)³. Differences of cubes can be factored as follows: a³ - b³ = (a - b)(a² + ab + b²).

Thus, (x²)³ - (y²)³ = (x² - y²)(x^4 + x²y² + y^4)