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Factorise :

a3b+a327b3a - 3b + a^3 - 27b^3

Factorisation

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Answer

(a3b)+(a327b3)=(a3b)+(a3b)(a2+a×3b+9b2)=(a3b)+(a3b)(a2+3ab+9b2)=(a3b)[1+(a2+3ab+9b2)]=(a3b)(1+a2+3ab+9b2)(a - 3b) + (a^3 - 27b^3) = (a - 3b) + (a - 3b)(a^2 + a \times 3b + 9b^2)\\[1em] = (a - 3b) + (a - 3b)(a^2 + 3ab + 9b^2)\\[1em] = (a - 3b)[1 + (a^2 + 3ab + 9b^2)]\\[1em] = (a - 3b)(1 + a^2 + 3ab + 9b^2)

Hence, a3b+a327b3=(a3b)(1+a2+3ab+9b2)a - 3b + a^3 - 27b^3 = (a - 3b)(1 + a^2 + 3ab + 9b^2).

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