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 $-3b=\frac{{\left(3\right)}^4}{3}=27$
$\Rightarrow b=-9$
$\therefore \left(a,b\right)=\left(3,-9\right)$