1999 CEMC Pascal Problems/Problem 2

Problem

If $k = 2$ , then $(k^3 - 8)(k + 1)$ equals

$\text{ (A) }\ 0 \qquad\text{ (B) }\ 3 \qquad\text{ (C) }\ 6 \qquad\text{ (D) }\ 8 \qquad\text{ (E) }\ -6$

Plugging in $k = 2$ , we get:

$(k^3 - 8)(k + 1) = (2^3 - 8)(2 + 1) = (8 - 8)(3) = 0 \times 3 = \boxed {\textbf {(A) } 0}$

~anabel.disher

Distributing the initial expression and then plugging in $k = 2$ , we have:

$(k^3 - 8)(k + 1) = k^4 - 8k + k^3 - 8$

$=2^4 - 8 \times 2 + 2^3 - 8$

$=16 - 8 \times 2 + 8 - 8$

$=16 - 16 + 8 - 8$

We can see that everything cancels out, giving that the expression is $\boxed {\textbf {(A) } 0}$ .

~anabel.disher