Question

KJ thought that (a-2)^3 is a^3-8. Why is that wrong?

Answer 1

The expression \( (a-2)^3 \) and \( a^3 - 8 \) are not the same, and here's why:

The expression \( (a-2)^3 \) represents the cube of the binomial \( (a-2) \). To expand \( (a-2)^3 \), we can use the binomial expansion or the formula for the cube of a binomial:

\[
(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3
\]

In this case, we have \( x = a \) and \( y = 2 \). Applying the formula:

\[
(a-2)^3 = a^3 - 3a^2(2) + 3a(2^2) - 2^3
\]

Calculating each term:

1. \( a^3 \)
2. \( - 3a^2(2) = -6a^2 \)
3. \( 3a(2^2) = 3a(4) = 12a \)
4. \( -2^3 = -8 \)

So, we can combine these results to get:

\[
(a-2)^3 = a^3 - 6a^2 + 12a - 8
\]

Now, let's consider the expression \( a^3 - 8 \). This expression can be factored as:

\[
a^3 - 8 = a^3 - 2^3 = (a-2)(a^2 + 2a + 4)
\]

Thus, \( a^3 - 8 \) is not equal to \( (a-2)^3 \) because the cubic expansion of \( (a-2) \) has additional terms (\(-6a^2 + 12a\)) that are not in \( a^3 - 8\).

In conclusion, the error in KJ's reasoning is that \( (a-2)^3 \) is not simply \( a^3 - 8 \); it involves additional terms from the expansion of the binomial, which makes the correct expansion equal to:

\[
(a-2)^3 = a^3 - 6a^2 + 12a - 8
\]

Hence, \( (a-2)^3 \neq a^3 - 8 \).