Answer :
To determine which expressions are sums of perfect cubes, we should identify if each expression can be rewritten as a sum of two perfect cubes.
Let's break down each expression one-by-one:
1. [tex]\(8x^6 + 27\)[/tex]:
- Here, [tex]\(8x^6 = (2x^2)^3\)[/tex] because [tex]\(8 = 2^3\)[/tex] and [tex]\(x^6 = (x^2)^3\)[/tex].
- Also, [tex]\(27 = 3^3\)[/tex].
- This expression is the sum of the cubes [tex]\((2x^2)^3 + 3^3\)[/tex].
2. [tex]\(x^9 + 1\)[/tex]:
- We can express [tex]\(x^9 = (x^3)^3\)[/tex] because raising a power to another power results in the product of the exponents.
- And [tex]\(1 = 1^3\)[/tex].
- This expression is the sum of the cubes [tex]\((x^3)^3 + 1^3\)[/tex].
3. [tex]\(81x^3 + 16x^6\)[/tex]:
- First, recognize [tex]\(81x^3 = (3x)^3\)[/tex] since [tex]\(81 = 3^4\)[/tex] but we only need to use [tex]\(3^3\)[/tex] for the cube.
- Next, [tex]\(16x^6 = (2x^2)^3\)[/tex] because [tex]\(16 = 2^4\)[/tex], but rearranged to match the cube, [tex]\(16 = 2^3\)[/tex].
- This expression can be rewritten as [tex]\((3x)^3 + (2x^2)^3\)[/tex].
4. [tex]\(x^6 + x^3\)[/tex]:
- Here, [tex]\(x^6 = (x^2)^3\)[/tex], as explained earlier.
- And [tex]\(x^3 = (x^1)^3\)[/tex], since we can assume [tex]\(x = x^1\)[/tex].
- This presents itself as [tex]\((x^2)^3 + (x)^3\)[/tex].
5. [tex]\(27x^9 + x^{12}\)[/tex]:
- [tex]\(27x^9 = (3x^3)^3\)[/tex] due to [tex]\(27 = 3^3\)[/tex].
- And [tex]\(x^{12} = (x^4)^3\)[/tex].
- Thus, it can be rewritten as [tex]\((3x^3)^3 + (x^4)^3\)[/tex].
6. [tex]\(9x^3 + 27x^9\)[/tex]:
- Notice [tex]\(9x^3 = (3x)^3\)[/tex] since [tex]\(9 = 3^2\)[/tex] needs adjustment but within cube context becomes [tex]\((3)^3\)[/tex].
- Similarly, [tex]\(27x^9 = (3x^3)^3\)[/tex] since [tex]\(27 = 3^3\)[/tex].
- This can be rewritten as [tex]\((3x)^3 + (3x^3)^3\)[/tex].
All these expressions can indeed be expressed as sums of perfect cubes, reaffirming that each one qualifies for the criteria provided.
Let's break down each expression one-by-one:
1. [tex]\(8x^6 + 27\)[/tex]:
- Here, [tex]\(8x^6 = (2x^2)^3\)[/tex] because [tex]\(8 = 2^3\)[/tex] and [tex]\(x^6 = (x^2)^3\)[/tex].
- Also, [tex]\(27 = 3^3\)[/tex].
- This expression is the sum of the cubes [tex]\((2x^2)^3 + 3^3\)[/tex].
2. [tex]\(x^9 + 1\)[/tex]:
- We can express [tex]\(x^9 = (x^3)^3\)[/tex] because raising a power to another power results in the product of the exponents.
- And [tex]\(1 = 1^3\)[/tex].
- This expression is the sum of the cubes [tex]\((x^3)^3 + 1^3\)[/tex].
3. [tex]\(81x^3 + 16x^6\)[/tex]:
- First, recognize [tex]\(81x^3 = (3x)^3\)[/tex] since [tex]\(81 = 3^4\)[/tex] but we only need to use [tex]\(3^3\)[/tex] for the cube.
- Next, [tex]\(16x^6 = (2x^2)^3\)[/tex] because [tex]\(16 = 2^4\)[/tex], but rearranged to match the cube, [tex]\(16 = 2^3\)[/tex].
- This expression can be rewritten as [tex]\((3x)^3 + (2x^2)^3\)[/tex].
4. [tex]\(x^6 + x^3\)[/tex]:
- Here, [tex]\(x^6 = (x^2)^3\)[/tex], as explained earlier.
- And [tex]\(x^3 = (x^1)^3\)[/tex], since we can assume [tex]\(x = x^1\)[/tex].
- This presents itself as [tex]\((x^2)^3 + (x)^3\)[/tex].
5. [tex]\(27x^9 + x^{12}\)[/tex]:
- [tex]\(27x^9 = (3x^3)^3\)[/tex] due to [tex]\(27 = 3^3\)[/tex].
- And [tex]\(x^{12} = (x^4)^3\)[/tex].
- Thus, it can be rewritten as [tex]\((3x^3)^3 + (x^4)^3\)[/tex].
6. [tex]\(9x^3 + 27x^9\)[/tex]:
- Notice [tex]\(9x^3 = (3x)^3\)[/tex] since [tex]\(9 = 3^2\)[/tex] needs adjustment but within cube context becomes [tex]\((3)^3\)[/tex].
- Similarly, [tex]\(27x^9 = (3x^3)^3\)[/tex] since [tex]\(27 = 3^3\)[/tex].
- This can be rewritten as [tex]\((3x)^3 + (3x^3)^3\)[/tex].
All these expressions can indeed be expressed as sums of perfect cubes, reaffirming that each one qualifies for the criteria provided.
