What is factor x^{12}+6x^8+12x^4+8 ?

The solution to factor x^{12}+6x^8+12x^4+8 is (x^4+2)^3

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factor x^{12}+6x^8+12x^4+8

factor

x^{12} + 6 x^{8} + 12 x^{4} + 8

(x^{4} + 2)^{3}

Solution steps

x^{12} + 6 x^{8} + 12 x^{4} + 8

Let

u = x^{2}

= u^{6} + 6 u^{4} + 12 u^{2} + 8

Factor

u^{6} + 6 u^{4} + 12 u^{2} + 8 : (u^{2} + 2) (u^{2} + 2) (u^{2} + 2)

= (u^{2} + 2) (u^{2} + 2) (u^{2} + 2)

Substitute back

u = x^{2}

= (x^{4} + 2) (x^{4} + 2) (x^{4} + 2)

Refine

= (x^{4} + 2)^{3}

(5 x + 1)^{2} + 15 (5 x + 1) + 56

2 cos^{2} (x) - 3 cos (x) - 2

(x + 3)^{3} + 8 x^{3}

e^{2 x} - 25

cos^{8} (x) - 2 cos^{4} (x) + 1

What is factor x^{12}+6x^8+12x^4+8 ?
The solution to factor x^{12}+6x^8+12x^4+8 is (x^4+2)^3