The sixty four times $x$ cube plus one is an expression in algebraic form and it should be factorized in the question. So, let’s learn how to factorise the algebraic expression $64$ times $x$ cube plus one in this problem.

First of all, let’s analyze the expression by studying the formation of each term in the given algebraic expression $64$ times $x$ cube plus $1$.

Two notable points can be observed in our analysis.

- A factor in algebraic form in the first term is in cube form.
- The two terms are connected by a plus sign.

The above two points are giving us a hint that the expression can be factorised, if each term is possible to convert in cube form. Now let’s learn how to convert each term in the expression in cube form.

Look at the first term. In this term, the first factor is a number, which is $64$ and the second factor is an expression in cube form. In order to express the first term in cube completely, the coefficient of $x$ cube should also be expressed in cube form. So, let’s factorize the number $64$ for expressing it in cube form.

$=\,\,$ $64 \times x^3+1$

$=\,\,$ $4 \times 16 \times x^3+1$

$=\,\,$ $4 \times 4 \times 4 \times x^3+1$

Now, the product of three identical numbers can be expressed in exponential notation in the first term.

$=\,\,$ $4^3 \times x^3+1$

The both factors in the first term of the expression are in cube form and their multiplication can be performed by the power rule of a product of the exponents.

$=\,\,$ $(4 \times x)^3+1$

$=\,\,$ $(4x)^3+1$

The first term is successfully converted into an expression in cube form. So, the second term should also be converted into cube form to transform the whole given expression as the sum of two cubes and the number one can be expressed in cube form by factorizing the number one as the product of three ones.

$=\,\,$ $(4x)^3+1 \times 1$

$=\,\,$ $(4x)^3+1 \times 1 \times 1$

$=\,\,$ $(4x)^3+1^3$

Therefore, the sixty-four times $x$ cube plus one is successfully converted as the sum of two cubes.

$\therefore\,\,\,\,$ $64x^3+1$ $\,=\,$ $(4x)^3+1^3$

The sum of cubes of four times $x$ and one can be factorized with the factorization by the sum of cubes method.

$a^3+b^3$ $\,=\,$ $(a+b)(a^2-ab+b^2)$

Let’s take $a \,=\, 4x$ and $b \,=\, 1$, and then let’s substitute them in the above factoring formula.

$\implies$ $(4x)^3+1^3$ $\,=\,$ $(4x+1)$ $\Big((4x)^2-(4x)(1)+1^2\Big)$

Now, let’s focus on simplifying each algebraic factor on the right-hand side of the algebraic expression.

$=\,\,\,$ $(4x+1)$ $\big(4^2 \times x^2$ $-$ $4x \times 1$ $+$ $1^2\big)$

$=\,\,\,$ $(4x+1)$ $\big(4 \times 4 \times x^2$ $-$ $4x \times 1$ $+$ $1 \times 1\big)$

$=\,\,\,$ $(4x+1)$ $\big(16 \times x^2$ $-$ $4x \times 1$ $+$ $1\big)$

$=\,\,\,$ $(4x+1)$ $(16x^2-4x+1)$

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