Question
The factors of $8a^3 + b^3 - 6ab + 1$ are:
✓
Answer
We know the identity
$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
So by using identity, we can write given expression as
$(2a)^3 + (b)^3 + (1)^3 - 3(2a)(b)(1)$
$= (2a + b + 1)[(2a)^2 + b^2 + 1^2 -2a \times b - b \times 1 - 2a \times 1]$
$= (2a + b + 1)(4a^2 + b^2 + 1 -2ab - b - 2a)$
$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
So by using identity, we can write given expression as
$(2a)^3 + (b)^3 + (1)^3 - 3(2a)(b)(1)$
$= (2a + b + 1)[(2a)^2 + b^2 + 1^2 -2a \times b - b \times 1 - 2a \times 1]$
$= (2a + b + 1)(4a^2 + b^2 + 1 -2ab - b - 2a)$
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