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(A) Given the expression $(1-x)^3(2+x)^6$. Since $x$ is very small,we neglect terms with $x^3$ and higher powers. \ $(1-x)^3 = 1 - 3x + 3x^2$. \ $(2

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-$80$

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(A) Given the expression $(1-x)^3(2+x)^6$. Since $x$ is very small,we neglect terms with $x^3$ and higher powers. \ $(1-x)^3 = 1 - 3x + 3x^2$. \ $(2+x)^6 = 2^6 + 6 	imes 2^5 	imes x + \frac{6 	imes 5}{2} 	imes 2^4 	imes x^2 = 64 + 192x + 240x^2$. \ Now,multiply the two expansions: \ $(1-3x+3x^2)(64+192x+240x^2) = 64 + 192x + 240x^2 - 192x - 576x^2 + 192x^2$ (neglecting $x^3$ and higher). \ $= 64 + (192-192)x + (240-576+192)x^2 = 64 + 0x - 144x^2$. \ Comparing with $a+bx+cx^2$,we get $a=64, b=0, c=-144$. \ Therefore,$a+b+c = 64 + 0 - 144 = -80$.

Let $x \in \mathbb{R}$ be so small that the powers of $x$ beyond two are insignificant and negligibly small. For such $x$,if $(1-x)^3(2+x)^6$ is approximated by $a+bx+cx^2$,then $a+b+c=$

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