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Multiply:
$x^2 + 4y^2 + 2xy - 3x + 6y + 9$ by $x - 2y + 3$
Given:
$x^2 + 4y^2 + 2xy - 3x + 6y + 9$ and $x - 2y + 3$
To do:
We have to multiply the given expressions.
Solution:
We know that,
$a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)$
Therefore,
$(x^2 + 4y^2 + 2xy - 3x + 6y + 9) \times (x - 2y + 3) = (x - 2y + 3) (x^2 + 4y^2 + 9 + 2xy + 6y - 3x)$
$= (x)^3 + (-2y)^3 + (3)^3 - 3 \times x \times (-2y) \times 3$
$= x^3 - 8y^3 + 27 + 18xy$
Hence, $(x^2 + 4y^2 + 2xy - 3x + 6y + 9) \times (x - 2y + 3) = x^3 - 8y^3 + 27 + 18xy$.
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