How to solve a ^ 60 - b ^ 30

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 ^ 60 - b ^ 30 | Equatix

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Question

$$a^{60}-b^{30}$$

Factor

$\left(a^{2}-b\right)\left(a^{2}+b\right)\left(a^{8}-a^{2}b^{3}+b^{4}+b^{2}a^{4}-ba^{6}\right)\left(a^{8}+a^{2}b^{3}+b^{4}+b^{2}a^{4}+ba^{6}\right)\left(a^{20}+b^{10}-b^{5}a^{10}\right)\left(a^{20}+b^{10}+b^{5}a^{10}\right)$

Solution Steps

Rewrite $a^{60}-b^{30}$ as $\left(a^{30}\right)^{2}-\left(b^{15}\right)^{2}$. The difference of squares can be factored using the rule: $p^{2}-q^{2}=\left(p-q\right)\left(p+q\right)$.

$$\left(a^{30}-b^{15}\right)\left(a^{30}+b^{15}\right)$$

Consider $a^{30}-b^{15}$. Rewrite $a^{30}-b^{15}$ as $\left(a^{10}\right)^{3}-\left(b^{5}\right)^{3}$. The difference of cubes can be factored using the rule: $p^{3}-q^{3}=\left(p-q\right)\left(p^{2}+pq+q^{2}\right)$.

$$\left(a^{10}-b^{5}\right)\left(a^{20}+b^{10}+b^{5}a^{10}\right)$$

Consider $a^{10}-b^{5}$. Consider $a^{10}-b^{5}$ as a polynomial over variable $a$. Find one factor of the form $a^{k}+m$, where $a^{k}$ divides the monomial with the highest power $a^{10}$ and $m$ divides the constant factor $-b^{5}$. One such factor is $a^{2}-b$. Factor the polynomial by dividing it by this factor.

$$\left(a^{2}-b\right)\left(a^{8}+a^{2}b^{3}+b^{4}+b^{2}a^{4}+ba^{6}\right)$$

Consider $a^{30}+b^{15}$. Rewrite $a^{30}+b^{15}$ as $\left(a^{10}\right)^{3}+\left(b^{5}\right)^{3}$. The sum of cubes can be factored using the rule: $p^{3}+q^{3}=\left(p+q\right)\left(p^{2}-pq+q^{2}\right)$.

$$\left(a^{10}+b^{5}\right)\left(a^{20}+b^{10}-b^{5}a^{10}\right)$$

Consider $a^{10}+b^{5}$. Consider $a^{10}+b^{5}$ as a polynomial over variable $a$. Find one factor of the form $a^{n}+u$, where $a^{n}$ divides the monomial with the highest power $a^{10}$ and $u$ divides the constant factor $b^{5}$. One such factor is $a^{2}+b$. Factor the polynomial by dividing it by this factor.

$$\left(a^{2}+b\right)\left(a^{8}-a^{2}b^{3}+b^{4}+b^{2}a^{4}-ba^{6}\right)$$

Rewrite the complete factored expression.

$$\left(a^{2}-b\right)\left(a^{2}+b\right)\left(a^{8}-a^{2}b^{3}+b^{4}+b^{2}a^{4}-ba^{6}\right)\left(a^{8}+a^{2}b^{3}+b^{4}+b^{2}a^{4}+ba^{6}\right)\left(a^{20}+b^{10}-b^{5}a^{10}\right)\left(a^{20}+b^{10}+b^{5}a^{10}\right)$$

Show Solution

Evaluate

$\left(a^{20}-b^{10}\right)\left(\left(a^{20}+b^{10}\right)^{2}-\left(b^{5}a^{10}\right)^{2}\right)$