Question

What is the factored form of x^12y^18+1?

Answer 1

Answer:

$(x^4y^6+1)(x^8y^{12}-x^4y^6+1)$

Step-by-step explanation:

We have been given an expression $x^{12}y^{18}+1$ and we are asked to factor our given expression.

We will use sum of cubes formula to factor our given expression.

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

Upon using power rule of exponents $a^{bc}=(a^b)^{c}$, we can write:

$x^{12}=(x^4)^3$

$y^{18}=(y^6)^3$

We can write 1 as $1^3$ as 1 raised to any power equals 1.

Now we are ready to apply sum of cubes formula.

$(x^4y^6)^3+1^3=(x^4y^6+1)((x^4y^6)^2-x^4y^6+1^2)$

Upon simplifying right side of our equation we will get,

$(x^4y^6)^3+1=(x^4y^6+1)(x^{4*2}y^{6*2}-x^4y^6+1)$

$(x^4y^6)^3+1=(x^4y^6+1)(x^8y^{12}-x^4y^6+1)$

Therefore, after factoring our given expression we get $(x^4y^6+1)(x^8y^{12}-x^4y^6+1)$.

Answer 2

The answer to that is (x^4 y^6+1)(x^8 y^12-x^4 y^6+1