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If $(a+b x)^{-3}=\frac{1}{27}+\frac{1}{3} x+\ldots$, then the ordered pair $(a, b)$ equals to
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Verified Answer
The correct answer is:
$(3,-9)$
$$
\begin{aligned}
\text { Now, }(a+b x)^{-3} & =a^{-3}\left(1+\frac{b x}{a}\right)^{-3} \\
& =\frac{1}{a^3}\left(1-{ }^3 C_1\left(\frac{b x}{a}\right)+\ldots\right) \\
\Rightarrow \quad \frac{1}{a^3} & -{ }^3 C_1 \frac{1}{a^3}\left(\frac{b x}{a}\right)+\ldots \\
& =\frac{1}{27}+\frac{1}{3} x
\end{aligned}
$$
(given)
On equating constant and $x$, we get
$$
\frac{1}{a^3}=\frac{1}{27}
$$
$\begin{array}{rlrl}\text { and } & -{ }^3 C_1 \frac{1}{a^3}\left(\frac{b}{a}\right) & =\frac{1}{3} \\ \Rightarrow & r l r l \\ \text { and } & -{ }^3 C_1\left(\frac{1}{3^4}\right) \times b & =\frac{1}{3} \\ \Rightarrow & -3 \times \frac{1}{3^4} b & =\frac{1}{3} \\ & \Rightarrow & -b & =3^2 \Rightarrow b=-9 \\ & \therefore & (a, b) & =(3,-9)\end{array}$
\begin{aligned}
\text { Now, }(a+b x)^{-3} & =a^{-3}\left(1+\frac{b x}{a}\right)^{-3} \\
& =\frac{1}{a^3}\left(1-{ }^3 C_1\left(\frac{b x}{a}\right)+\ldots\right) \\
\Rightarrow \quad \frac{1}{a^3} & -{ }^3 C_1 \frac{1}{a^3}\left(\frac{b x}{a}\right)+\ldots \\
& =\frac{1}{27}+\frac{1}{3} x
\end{aligned}
$$
(given)
On equating constant and $x$, we get
$$
\frac{1}{a^3}=\frac{1}{27}
$$
$\begin{array}{rlrl}\text { and } & -{ }^3 C_1 \frac{1}{a^3}\left(\frac{b}{a}\right) & =\frac{1}{3} \\ \Rightarrow & r l r l \\ \text { and } & -{ }^3 C_1\left(\frac{1}{3^4}\right) \times b & =\frac{1}{3} \\ \Rightarrow & -3 \times \frac{1}{3^4} b & =\frac{1}{3} \\ & \Rightarrow & -b & =3^2 \Rightarrow b=-9 \\ & \therefore & (a, b) & =(3,-9)\end{array}$
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