Question

Factorise: a3-8b3 + 1 + 6ab​

Answer 1

Step-by-step explanation:

a³-8b³+1+6ab

=(a)³+(-2b)³+(1)³-3(a)(-2b)(1)

=[a+(-2b)+1][a²+(-2b)²+(1)³-(a)(-2b)-(-2b)(1)-(1)(a)]

=(a-2b+1)(a²+4b²+1+2ab+2b-a)

Answer 2

The required factorization is $(a-2b+1)(a^2+4b^2+1+2ab+2b-a)$.

Step-by-step explanation:

Consider the provided expression.

$a^3-8b^3 + 1 + 6ab$

Use the identity: $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$

Now rewrite the provided expression.

$(a)^3-(2b)^3 + (1)^3 - 3(a)(-2b)(1)$

Where x=a, y=-2b, and z=1

Therefore, the required expression is:

$(a-2b+1)(a^2+4b^2+1+2ab+2b-a)$

Hence, the required factorization is $(a-2b+1)(a^2+4b^2+1+2ab+2b-a)$.

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