What is factor 3a^{12}+1+3a^6+a^{18} ?

The solution to factor 3a^{12}+1+3a^6+a^{18} is (a^2+1)^3(a^{12}-3a^{10}+6a^8-7a^6+6a^4-3a^2+1)

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factor 3a^{12}+1+3a^6+a^{18}

factor

3 a^{12} + 1 + 3 a^{6} + a^{18}

(a^{2} + 1)^{3} (a^{12} - 3 a^{10} + 6 a^{8} - 7 a^{6} + 6 a^{4} - 3 a^{2} + 1)

Solution steps

3 a^{12} + 1 + 3 a^{6} + a^{18}

Write in the standard form

= a^{18} + 3 a^{12} + 3 a^{6} + 1

Let

u = a^{2}

= u^{9} + 3 u^{6} + 3 u^{3} + 1

Factor

u^{9} + 3 u^{6} + 3 u^{3} + 1 : (u + 1) (u + 1) (u + 1) (u^{6} - 3 u^{5} + 6 u^{4} - 7 u^{3} + 6 u^{2} - 3 u + 1)

= (u + 1) (u + 1) (u + 1) (u^{6} - 3 u^{5} + 6 u^{4} - 7 u^{3} + 6 u^{2} - 3 u + 1)

Substitute back

u = a^{2}

= (a^{2} + 1) (a^{2} + 1) (a^{2} + 1) (a^{12} - 3 a^{10} + 6 a^{8} - 7 a^{6} + 6 a^{4} - 3 a^{2} + 1)

Apply exponent rule:

aaa = a^{3}

(a^{2} + 1) (a^{2} + 1) (a^{2} + 1) = (a^{2} + 1)^{3}

= (a^{2} + 1)^{3} (a^{12} - 3 a^{10} + 6 a^{8} - 7 a^{6} + 6 a^{4} - 3 a^{2} + 1)

(6 + x)^{3}

(a + b)^{3} + 1

x (x + y)^{3} - 3 x^{2} y (x + y)

9 x^{2} + 12 x + 4 - 16 y^{2}

2 sin^{2} (a) + 2 cos^{2} (a)

What is factor 3a^{12}+1+3a^6+a^{18} ?
The solution to factor 3a^{12}+1+3a^6+a^{18} is (a^2+1)^3(a^{12}-3a^{10}+6a^8-7a^6+6a^4-3a^2+1)