How to solve x ^ 12 - y ^ 12

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve x ^ 12 - y ^ 12 | Equatix

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Question

$$x^{12}-y^{12}$$

Factor

$\left(x-y\right)\left(x+y\right)\left(x^{2}+y^{2}\right)\left(x^{2}-xy+y^{2}\right)\left(x^{2}+xy+y^{2}\right)\left(x^{4}-x^{2}y^{2}+y^{4}\right)$

Solution Steps

Rewrite $x^{12}-y^{12}$ as $\left(x^{6}\right)^{2}-\left(y^{6}\right)^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.

$$\left(x^{6}-y^{6}\right)\left(x^{6}+y^{6}\right)$$

Consider $x^{6}-y^{6}$. Rewrite $x^{6}-y^{6}$ as $\left(x^{3}\right)^{2}-\left(y^{3}\right)^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.

$$\left(x^{3}-y^{3}\right)\left(x^{3}+y^{3}\right)$$

Consider $x^{3}-y^{3}$. The difference of cubes can be factored using the rule: $a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right)$.

$$\left(x-y\right)\left(x^{2}+xy+y^{2}\right)$$

Consider $x^{3}+y^{3}$. The sum of cubes can be factored using the rule: $a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right)$.

$$\left(x+y\right)\left(x^{2}-xy+y^{2}\right)$$

Consider $x^{6}+y^{6}$. Rewrite $x^{6}+y^{6}$ as $\left(x^{2}\right)^{3}+\left(y^{2}\right)^{3}$. The sum of cubes can be factored using the rule: $a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right)$.

$$\left(x^{2}+y^{2}\right)\left(x^{4}-x^{2}y^{2}+y^{4}\right)$$

Rewrite the complete factored expression.

$$\left(x-y\right)\left(x+y\right)\left(x^{2}-xy+y^{2}\right)\left(x^{2}+xy+y^{2}\right)\left(x^{4}-x^{2}y^{2}+y^{4}\right)\left(x^{2}+y^{2}\right)$$

Show Solution

Evaluate

$\left(x^{4}-y^{4}\right)\left(x^{4}+y^{4}-\left(xy\right)^{2}\right)\left(-\left(xy\right)^{2}+\left(x^{2}+y^{2}\right)^{2}\right)$