Question

If $a=(2^{-2}-2^{-3}),b=(2^{-3}-2^{-4})$ and $c=(2^{-4}-2^{-2})$, then $a^3+b^3+c^3$

• A. $\frac{-9}{1024}$
• B. $\frac{-9}{2048}$
• C. $0$
• D. $1$

Answer 1

The correct option is B $\frac{-9}{2048}$

$a=2^{-2}-2^{-3}=\frac{1}{4}-\frac{1}{3}=\frac{1}{8}=\frac{2}{16}$

$b=2^{-2}-2^{-4}=\frac{1}{8}-\frac{1}{16}=\frac{1}{16}$

$c=2^{-4}-2^{-2}=\frac{1}{16}-\frac{4}{16}=\frac{-3}{16}$

$a^3+b^3+C^3=\frac{8}{{16}^3}+\frac{1}{{16}^3}-\frac{27}{{16}^3}$

$=\frac{-18}{{16}^3}=\frac{-9}{2048}$