Question
Multiply: $\left(x^2+4 y^2+2 x y-3 x+6 y+9\right)$ by $(x-2 y+3)$
✓
Answer
$=(x-2 y+3)\left(x^2+4 y^2+9+2 x y+6 y-3 x\right)$
$=(x+(-2 y)+3)\left(x^2+(-2 y)^2+3^2-x(-2 y)-(-2 y) 3-3 x\right)$
$=x^3+(-2 y)^3+3^3-3 \times x(-2 y)^3\left[\because(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)=a^3+b^3+c^3-3 a b c\right]$
$=x^3-8 y^3+27+18 x y$
$=(x+(-2 y)+3)\left(x^2+(-2 y)^2+3^2-x(-2 y)-(-2 y) 3-3 x\right)$
$=x^3+(-2 y)^3+3^3-3 \times x(-2 y)^3\left[\because(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)=a^3+b^3+c^3-3 a b c\right]$
$=x^3-8 y^3+27+18 x y$
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