Q.128.Factorization

Question 128.
Factorize (9x^3 - 1)^3 + (9x^4 - 3x)^3.

Answer 128.
(9x^3 - 1)^3 = 729x^9 - 243x^6 + 27x^3 - 1 ... (1)
(9x^4 - 3x)^3 = 729x^12 - 729x^9 + 243x^6 - 27x^3 ... (2)

Adding (1) and (2),
(9*x^3 - 1)^3 + (9*x^4 - 3*x)^3
= 729x^9 - 243x^6 + 27x^3 - 1 + 729x^12 - 729x^9 + 243x^6 - 27x^3
= 729x^12 - 1
= (9x^4)^3 - 1^3
= (9x^4 - 1) (81x^8 + 9x^4 + 1)
= (3x^2 - 1)(3x^2 + 1)[(81x^8 + 18x^4 + 1) - 9x^4]
= (3x^2 - 1)(3x^2 + 1)[(9x^4 + 1)^2 - (3x^2)^2]
= (3x^2 - 1)(3x^2 + 1)(9x^4 + 3x^2 + 1)(9x^4 - 3x^2 + 1)
= (3x^2 - 1)(3x^2 + 1)(9x^4 + 3x^2 + 1)[(9x^4 + 6x^2 + 1) - 9x^2]
= (3x^2 - 1)(3x^2 + 1)(9x^4 + 3x^2 + 1)[(3x^2 + 1)^2 - (3x)^2]
= (3x^2 - 1)(3x^2 + 1)(9x^4 + 3x^2 + 1)(3x^2 + 3x + 1)(3x^2-3x+1).

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