Question

The value of $\frac{(1 - d^3)}{(1 - d)}$ is:

• A. > 1$ if $d>-1$
• B. > 3$ if $d>1$
• C. > 2$ if $0<d<0.5$
• D. $< 2$ if $d<-2$

Hint:

When a cubic minus a constant appears in a fraction, factor using the $a^3 - b^3$ identity to simplify and check inequalities.

Answer 1

The Correct Option is B

We start with:
$\frac{1 - d^3}{1 - d}$. Using the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, we have:
$1 - d^3 = (1 - d)(1 + d + d^2)$.
Thus:
$\frac{1 - d^3}{1 - d} = 1 + d + d^2$.
Now, if $d>1$, then $d^2>d>1$, so:
$1 + d + d^2>1 + 1 + 1 = 3$.
In fact, since $d>1$, $d^2$ is strictly greater than $1$, so the sum is strictly greater than $3$.
Thus, the condition "> 3$ if $d>1$" is correct.