Question

think about the polynomial identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$, for any values of $a$ and $b$ where $a + b$ is not 0 or 1. what is true about the sum of $a^3 + b^3$ for any values of $a$ and $b$ where $a + b$ is not 0 or 1? \
\
$\bigcirc$ the sum is a composite number. \
$\bigcirc$ the sum is equal to 0. \
$\bigcirc$ the sum is a prime number. \
$\bigcirc$ the sum is equal to 1.

Explanation:

Step1: Recall the sum of cubes formula

The given identity is \(a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})\). We need to analyze the sum \(a^{3}+b^{3}\) when \(a + b
eq0\) and \(a + b
eq1\).

Step2: Analyze the factors

Since \(a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})\), and \(a + b\) is a factor (and \(a + b
eq0,1\)), let's consider the case when \(a + b\) is an integer (for simplicity, though the concept holds generally). A composite number is a positive integer that has at least one positive divisor other than one or itself. If \(a + b\) is an integer greater than 1 (since \(a + b
eq0,1\)) and \(a^{2}-ab + b^{2}\) is also an integer (for integer \(a,b\)), then \(a^{3}+b^{3}\) can be factored into two integers greater than 1 (if \(a^{2}-ab + b^{2}>1\) and \(a + b>1\)). Let's take an example: let \(a = 2\), \(b = 3\). Then \(a + b=5\), \(a^{2}-ab + b^{2}=4 - 6+9 = 7\), and \(a^{3}+b^{3}=8 + 27=35=5\times7\), which is composite. If \(a=1\), \(b = 2\), \(a + b = 3\), \(a^{2}-ab + b^{2}=1-2 + 4 = 3\), \(a^{3}+b^{3}=1 + 8 = 9=3\times3\), composite. Now check the other options: The sum is not always 0 (since \(a + b
eq0\), and if \(a + b
eq0\), \(a^{3}+b^{3}\) won't be 0 unless \(a=b = 0\) but \(a + b
eq0\) here). The sum is not always 1 (e.g., \(a=2,b=3\) gives 35). A prime number has only 1 and itself as factors, but \(a^{3}+b^{3}\) can be factored into \((a + b)\) and \((a^{2}-ab + b^{2})\), so it's composite (when factors are greater than 1).

Answer:

The sum is a composite number.