What is factor a^{20}-a^{16}+a^{12}-a^8+a^4-a^2 ?

The solution to factor a^{20}-a^{16}+a^{12}-a^8+a^4-a^2 is a^2(a+1)(a-1)(a^{16}+a^{14}+a^8+a^6+1)

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factor a^{20}-a^{16}+a^{12}-a^8+a^4-a^2

factor

a^{20} - a^{16} + a^{12} - a^{8} + a^{4} - a^{2}

a^{2} (a + 1) (a - 1) (a^{16} + a^{14} + a^{8} + a^{6} + 1)

Solution steps

a^{20} - a^{16} + a^{12} - a^{8} + a^{4} - a^{2}

Factor out common term

a^{2} : a^{2} (a^{18} - a^{14} + a^{10} - a^{6} + a^{2} - 1)

= a^{2} (a^{18} - a^{14} + a^{10} - a^{6} + a^{2} - 1)

Factor

a^{18} - a^{14} + a^{10} - a^{6} + a^{2} - 1 : (a + 1) (a - 1) (a^{16} + a^{14} + a^{8} + a^{6} + 1)

= a^{2} (a + 1) (a - 1) (a^{16} + a^{14} + a^{8} + a^{6} + 1)

2 (x + 12) = 3 x + 24 - x

4 x^{2} + 2 x + 2 = 0

x + 4 = 3 x + 12

2 (5 y + 2) = 14

3 x - x + 10 + 6 x = 18

What is factor a^{20}-a^{16}+a^{12}-a^8+a^4-a^2 ?
The solution to factor a^{20}-a^{16}+a^{12}-a^8+a^4-a^2 is a^2(a+1)(a-1)(a^{16}+a^{14}+a^8+a^6+1)