Answer :
Sure, let's factor [tex]\(64x^6 - 27\)[/tex] using the identity for the difference of cubes:
[tex]\[a^3 - b^3 = (a - b)\left(a^2 + ab + b^2\right).\][/tex]
First, we need to express the given expression [tex]\(64x^6 - 27\)[/tex] in terms of cubes. Let’s identify [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that [tex]\(a^3 = 64x^6\)[/tex] and [tex]\(b^3 = 27\)[/tex].
Now, let's break it down step-by-step:
1. For [tex]\(64x^6\)[/tex] to be a cube, we note that [tex]\(64 = 4^3\)[/tex] and [tex]\(x^6 = (x^2)^3\)[/tex]. Therefore, we can write:
[tex]\[64x^6 = (4x^2)^3.\][/tex]
So, [tex]\(a = 4x^2\)[/tex].
2. For 27 to be a cube, we recognize that:
[tex]\[27 = 3^3.\][/tex]
So, [tex]\(b = 3\)[/tex].
Now, we have identified [tex]\(a\)[/tex] and [tex]\(b\)[/tex] as follows:
[tex]\[ \begin{array}{l} a = 4x^2 \\ b = 3 \end{array} \][/tex]
Hence, the expression [tex]\(64x^6 - 27\)[/tex] can be rewritten using the difference of cubes formula as:
[tex]\[64x^6 - 27 = (4x^2)^3 - 3^3 = (4x^2 - 3)\left((4x^2)^2 + (4x^2)(3) + 3^2\right).\][/tex]
[tex]\[a^3 - b^3 = (a - b)\left(a^2 + ab + b^2\right).\][/tex]
First, we need to express the given expression [tex]\(64x^6 - 27\)[/tex] in terms of cubes. Let’s identify [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that [tex]\(a^3 = 64x^6\)[/tex] and [tex]\(b^3 = 27\)[/tex].
Now, let's break it down step-by-step:
1. For [tex]\(64x^6\)[/tex] to be a cube, we note that [tex]\(64 = 4^3\)[/tex] and [tex]\(x^6 = (x^2)^3\)[/tex]. Therefore, we can write:
[tex]\[64x^6 = (4x^2)^3.\][/tex]
So, [tex]\(a = 4x^2\)[/tex].
2. For 27 to be a cube, we recognize that:
[tex]\[27 = 3^3.\][/tex]
So, [tex]\(b = 3\)[/tex].
Now, we have identified [tex]\(a\)[/tex] and [tex]\(b\)[/tex] as follows:
[tex]\[ \begin{array}{l} a = 4x^2 \\ b = 3 \end{array} \][/tex]
Hence, the expression [tex]\(64x^6 - 27\)[/tex] can be rewritten using the difference of cubes formula as:
[tex]\[64x^6 - 27 = (4x^2)^3 - 3^3 = (4x^2 - 3)\left((4x^2)^2 + (4x^2)(3) + 3^2\right).\][/tex]
