Solve (a^2-2a-1)(a^2-2a+1)

$$( a ^ { 2 } - 2 a - 1 ) ( a ^ { 2 } - 2 a + 1 )$$

Evaluate

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

Solution Steps

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}, where a=a^{2}-2a and b=1. Square 1.

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

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a^{2}-2a\right)^{2}.

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

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

$$a^{4}-4a^{2}a+4a^{2}-1$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$a^{4}-4a^{3}+4a^{2}-1$$

Expand

$$a^{4}-4a^{3}+4a^{2}-1$$

Solution Steps

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}, where a=a^{2}-2a and b=1. Square 1.

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

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a^{2}-2a\right)^{2}.

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

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

$$a^{4}-4a^{2}a+4a^{2}-1$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$a^{4}-4a^{3}+4a^{2}-1$$