Solve (a^{2})^{2}-(a^2)^4*(a^3)^2

$$( a ^ { 2 } ) ^ { 2 } - ( a ^ { 2 } ) ^ { 4 } \cdot ( a ^ { 3 } ) ^ { 2 }$$

Evaluate

$$a^{4}-a^{14}$$

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$a^{4}-\left(a^{2}\right)^{4}\left(a^{3}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 4 to get 8.

$$a^{4}-a^{8}\left(a^{3}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$a^{4}-a^{8}a^{6}$$

To multiply powers of the same base, add their exponents. Add 8 and 6 to get 14.

$$a^{4}-a^{14}$$

Factor

$$\left(a-1\right)\left(a+1\right)a^{4}\left(-a^{4}+a^{3}-a^{2}+a-1\right)\left(a^{4}+a^{3}+a^{2}+a+1\right)$$

Solution Steps

Factor out common term a^{4} by using distributive property.

$$a^{4}\left(1-a^{10}\right)$$

Consider 1-a^{10}. Rewrite 1-a^{10} as 1^{2}-\left(-a^{5}\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(1+a^{5}\right)\left(1-a^{5}\right)$$

Reorder the terms.

$$\left(a^{5}+1\right)\left(-a^{5}+1\right)$$

Consider a^{5}+1. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient 1. One such root is -1. Factor the polynomial by dividing it by a+1.

$$\left(a+1\right)\left(a^{4}-a^{3}+a^{2}-a+1\right)$$

Consider -a^{5}+1. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient -1. One such root is 1. Factor the polynomial by dividing it by a-1.

$$\left(a-1\right)\left(-a^{4}-a^{3}-a^{2}-a-1\right)$$

Rewrite the complete factored expression.

$$a^{4}\left(-a^{4}-a^{3}-a^{2}-a-1\right)\left(a-1\right)\left(a^{4}-a^{3}+a^{2}-a+1\right)\left(a+1\right)$$