Answer :

We start with the expression

[tex]$$
64c\,z^3 + 27c\,x^3.
$$[/tex]

Step 1. Factor out the common factor [tex]$c$[/tex]:

Factor [tex]$c$[/tex] from the expression:

[tex]$$
64c\,z^3 + 27c\,x^3 = c\left(64z^3 + 27x^3\right).
$$[/tex]

Step 2. Recognize the sum of cubes:

Notice that
- [tex]$64z^3 = (4z)^3$[/tex], and
- [tex]$27x^3 = (3x)^3$[/tex].

Thus, the expression inside the parentheses becomes a sum of two cubes:

[tex]$$
64z^3 + 27x^3 = (4z)^3 + (3x)^3.
$$[/tex]

Step 3. Apply the sum of cubes formula:

Recall the sum of cubes formula:

[tex]$$
a^3 + b^3 = (a + b)(a^2 - ab + b^2).
$$[/tex]

Here, we set
[tex]$$
a = 4z \quad \text{and} \quad b = 3x.
$$[/tex]

Substitute these into the formula:

[tex]$$
(4z)^3 + (3x)^3 = (4z + 3x)\left[(4z)^2 - (4z)(3x) + (3x)^2\right].
$$[/tex]

Step 4. Simplify the factors:

Calculate each term in the second factor:
- [tex]$(4z)^2 = 16z^2$[/tex],
- [tex]$(3x)^2 = 9x^2$[/tex], and
- [tex]$(4z)(3x) = 12zx$[/tex].

Thus,

[tex]$$
(4z)^2 - (4z)(3x) + (3x)^2 = 16z^2 - 12zx + 9x^2.
$$[/tex]

Step 5. Write the complete factorization:

Substitute back into the factored expression:

[tex]$$
64c\,z^3 + 27c\,x^3 = c \, (4z + 3x)\, \left(16z^2 - 12zx + 9x^2\right).
$$[/tex]

For clarity, you may also write the binomial in the form [tex]$(3x + 4z)$[/tex], so the answer is equivalently:

[tex]$$
c \, (3x + 4z)\, \left(9x^2 - 12xz + 16z^2\right).
$$[/tex]

This is the fully factored form of the original expression.

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Rewritten by : Barada