Question

Someone please help me

Answer 1

Step-by-step explanation:

Part A: $u^6$ can be written as the square of u³, or $(u^3)^2$. Similarly, $v^6=(v^3)^2$. Hence, we can write this as a difference of two squares by writing it as

$(u^3)^2-(v^3)^2$

Part B:

Difference of Two Squares

We can first factor a difference of two squares a² - b² into (a+b)(a-b). Here, a would be u³ and b would be v³.

$(u^3+v^3)(u^3-v^3)$

Sum and Difference of Two Cubes

We can factor this further by the use of two special formulas to factor a sum of two cubes and a difference of two cubes. These formulas are as follows:

$a^3+b^3=(a+b)(a^2-ab+b^2)\\a^3-b^3=(a-b)(a^2+ab+b^2)$

Since u³ + v³ is a sum of two cubes, let's rewrite it.

$u^3+v^3=(u+v)(u^2-uv+v^2)$

Since u³ - v³ is a difference of two cubes, we can rewrite it as well.

$u^3-v^3=(u-v)(u^2+uv+v^2)$

Now, let's multiply them together again to get the final factored form.

$u^6-v^6=(u+v)(u^2-uv+v^2)(u-v)(u^2+uv+v^2)$

Part C:

If we want to factor $x^6-1$ completely, we can just see that x to the sixth power is just $x^6$ and 1 to the sixth power is just 1. Hence, x can substitute for u and 1 can substitute for v.

$x^6-1=(x+1)(x^2-x(1)+1^2)(x-1)(x^2+x(1)+1^2)\\x^6-1=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$

We can repeat this for $x^6-64$, as 64 is just 2 to the sixth power.

$x^6-64=(x+2)(x^2-x(2)+2^2)(x-2)(x^2+x(2)+2^2)\\x^6-64=(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)$