Question

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

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

Answer 1

Answer: (D)

$\left(a+bx\right)^{-3} = \frac{1}{27}+\frac{1}{3} x + ....$
Now $\left(a+bx\right)^{-3} = a^{-3}\left(1+\frac{bx}{a}\right)^{-3}$
$= \frac{1}{a^{3}}\left(1-3 \frac{bx}{a}+\frac{\left(-3\right)\left(-4\right)}{2} \frac{b^{2}x^{2}}{a^{2}}....\right)$
$\therefore \frac{1}{27} = \frac{1}{a^{3} }$ and $\frac{-3b}{a^{4}} = \frac{1}{3}$
$\Rightarrow a^{3} = 27$ i.e., $a = 3$ and $-3b = \frac{\left(3\right)^{4}}{3} = 27$
$\Rightarrow b = -9$
$\therefore \left(a, b\right) = \left(3, -9\right)$