How to solve a ^ 6 - b ^ 6

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 a ^ 6 - b ^ 6 | Equatix

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Question

$$a^{6}-b^{6}$$

Factor

$\left(a-b\right)\left(a+b\right)\left(a^{2}-ab+b^{2}\right)\left(a^{2}+ab+b^{2}\right)$

Solution Steps

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

$$\left(a^{3}-b^{3}\right)\left(a^{3}+b^{3}\right)$$

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

$$\left(a-b\right)\left(a^{2}+ab+b^{2}\right)$$

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

$$\left(a+b\right)\left(a^{2}-ab+b^{2}\right)$$

Rewrite the complete factored expression.

$$\left(a-b\right)\left(a+b\right)\left(a^{2}-ab+b^{2}\right)\left(a^{2}+ab+b^{2}\right)$$

Show Solution

Evaluate

$\left(a^{2}-b^{2}\right)\left(-\left(ab\right)^{2}+\left(a^{2}+b^{2}\right)^{2}\right)$