Find the value


Find the value

$x^{3}+6 x^{2}+12 x+16$


$=x^{3}+6 x^{2}+12 x+8+8$

$=x^{3}+3 \times x^{2} \times 2+3 \times x \times 2^{2}+2^{3}+8$


$\left[\therefore a^{3}+3 a^{2} b+3 a b^{2}+b^{3}=(a+b)^{3}\right]$



$\therefore\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$

$=(x+2+2)\left(x^{2}+4+4 x-2 x-4+4\right)$

$\left[\therefore(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$=(x+4)\left(x^{2}+4+2 x\right)$

$\therefore x^{3}+6 x^{2}+12 x+16=(x+4)\left(x^{2}+4+2 x\right)$

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