Solve a^120-b^120

$$a ^ { 120 } - b ^ { 120 }$$

Factor

$$\left(a-b\right)\left(a+b\right)\left(a^{2}+b^{2}\right)\left(a^{4}+b^{4}\right)\left(a^{4}+a^{2}b^{2}-ab^{3}+b^{4}-ba^{3}\right)\left(a^{4}+a^{2}b^{2}+ab^{3}+b^{4}+ba^{3}\right)\left(a^{8}+a^{4}b^{4}-a^{2}b^{6}+b^{8}-b^{2}a^{6}\right)\left(a^{10}-a^{5}b^{5}+b^{10}\right)\left(a^{10}+a^{5}b^{5}+b^{10}\right)\left(a^{16}+a^{8}b^{8}-a^{4}b^{12}+b^{16}-b^{4}a^{12}\right)\left(a^{20}-a^{10}b^{10}+b^{20}\right)\left(a^{40}-a^{20}b^{20}+b^{40}\right)$$

Solution Steps

Rewrite a^{120}-b^{120} as \left(a^{60}\right)^{2}-\left(b^{60}\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(a^{60}-b^{60}\right)\left(a^{60}+b^{60}\right)$$

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

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

Consider a^{30}-b^{30}. Rewrite a^{30}-b^{30} as \left(a^{15}\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).

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

Consider a^{15}-b^{15}. Rewrite a^{15}-b^{15} as \left(a^{5}\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).

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

Consider a^{5}-b^{5}. Consider a^{5}-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^{5} and m divides the constant factor -b^{5}. One such factor is a-b. Factor the polynomial by dividing it by this factor.

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

Consider a^{15}+b^{15}. Rewrite a^{15}+b^{15} as \left(a^{5}\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).

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

Consider a^{5}+b^{5}. Consider a^{5}+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^{5} and u divides the constant factor b^{5}. One such factor is a+b. Factor the polynomial by dividing it by this factor.

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

Consider a^{30}+b^{30}. Rewrite a^{30}+b^{30} as \left(a^{10}\right)^{3}+\left(b^{10}\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).

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

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

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

Consider a^{60}+b^{60}. Rewrite a^{60}+b^{60} as \left(a^{20}\right)^{3}+\left(b^{20}\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).

$$\left(a^{20}+b^{20}\right)\left(a^{40}-a^{20}b^{20}+b^{40}\right)$$

Consider a^{20}+b^{20}. Consider a^{20}+b^{20} as a polynomial over variable a. Find one factor of the form a^{c}+d, where a^{c} divides the monomial with the highest power a^{20} and d divides the constant factor b^{20}. One such factor is a^{4}+b^{4}. Factor the polynomial by dividing it by this factor.

$$\left(a^{4}+b^{4}\right)\left(a^{16}+a^{8}b^{8}-a^{4}b^{12}+b^{16}-b^{4}a^{12}\right)$$

Rewrite the complete factored expression.

$$\left(a-b\right)\left(a+b\right)\left(a^{4}+a^{2}b^{2}-ab^{3}+b^{4}-ba^{3}\right)\left(a^{4}+a^{2}b^{2}+ab^{3}+b^{4}+ba^{3}\right)\left(a^{8}+a^{4}b^{4}-a^{2}b^{6}+b^{8}-b^{2}a^{6}\right)\left(a^{16}+a^{8}b^{8}-a^{4}b^{12}+b^{16}-b^{4}a^{12}\right)\left(a^{10}-a^{5}b^{5}+b^{10}\right)\left(a^{10}+a^{5}b^{5}+b^{10}\right)\left(a^{20}-a^{10}b^{10}+b^{20}\right)\left(a^{40}-a^{20}b^{20}+b^{40}\right)\left(a^{2}+b^{2}\right)\left(a^{4}+b^{4}\right)$$

Evaluate

$$\left(a^{20}+b^{20}-\left(ab\right)^{10}\right)\left(-\left(ab\right)^{10}+\left(a^{10}+b^{10}\right)^{2}\right)\left(a^{40}-b^{40}\right)\left(a^{40}+b^{40}-\left(ab\right)^{20}\right)$$