If (a+bx)^{-3} = frac{1}{27} + frac{1}{3}x + | vedclass

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(D) Given the expansion: $(a+bx)^{-3} = \frac{1}{27} + \frac{1}{3}x + \dots$ We can write $(a+bx)^{-3} = a^{-3}(1 + \frac{bx}{a})^{-3}$. Using the bin

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$(3, -9)$

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(D) Given the expansion: $(a+bx)^{-3} = \frac{1}{27} + \frac{1}{3}x + \dots$ We can write $(a+bx)^{-3} = a^{-3}(1 + \frac{bx}{a})^{-3}$. Using the binomial expansion $(1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \dots$,we get: $a^{-3}(1 + (-3)(\frac{bx}{a}) + \dots) = \frac{1}{a^3} - \frac{3bx}{a^4} + \dots$ Comparing the constant terms: $\frac{1}{a^3} = \frac{1}{27} \implies a^3 = 27 \implies a = 3$. Comparing the coefficients of $x$: $-\frac{3b}{a^4} = \frac{1}{3}$. Substituting $a=3$: $-\frac{3b}{3^4} = \frac{1}{3} \implies -\frac{b}{3^3} = \frac{1}{3} \implies -\frac{b}{27} = \frac{1}{3} \implies b = -\frac{27}{3} = -9$. Thus,the ordered pair $(a, b) = (3, -9)$.

If $(a+bx)^{-3} = \frac{1}{27} + \frac{1}{3}x + \dots$,then the ordered pair $(a, b)$ is equal to

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