Question

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

• A. $(3,-27)$
• B. $(1,\frac{1}{3})$
• C. $(3,9)$
• D. $(3,-9)$

Answer 1

The correct option is D $(3,-9)$

$(a+bx{)}^{-3}=\frac{1}{a^3}{(1+\frac{bx}{a})}^{-3}=\frac{1}{a^3}[1-\frac{3bx}{a}+...]$

$=\frac{1}{27}+\frac{x}{3}+...$..........(given)

On comparing, we get

$\frac{1}{a^3}=\frac{1}{27}$

$\Rightarrow a=3$

$\frac{x}{3}=\frac{-3bx}{a^4}$

$\Rightarrow b=-9$

$\therefore (a,b)=(3,-9)$