Question

How do you factor $64{x^3} - 27?$

Answer 1

Here we note that the terms in the given equation are in the form of perfect cubes of some numbers. Hence, we use the basic algebraic formula to solve the above equation.
In this problem we factorize the given equation using the difference of cubes formula which is given by ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$. But in the given equation we have additionally coefficient for the variable and a constant, so we need to factorize the coefficient of the given variable, constant at the same time and we will convert the given equation in the form of ${a^3} - {b^3}$, by applying some exponential rules and then we will use the algebraic formula and write the given equation as the product of its factors.
Note that the obtained result contains a linear equation as one term and quadratic equation as the other term.

Complete step by step solution:
Given the equation $64{x^3} - 27$.
Consider the coefficient of ${x^3}$, which is 64.
Now we factorize 64.
Dividing 64 with 2, we will get zero as remainder and 32 as quotient.
So, we can write 64 as,
$64 = 2 \times 32.$
Now we will factorize 32. We know that the 32 is divisible by 2. So, we will get zero as remainder and 16 as quotient.
So, we can write 32 as,
$32 = 2 \times 16.$
Now we will factorize 16. We know that 16 can be written as $16 = 4 \times 4.$
From all the above values, the value of 64 can be written as,
$64 = 2 \times 32$
$\Rightarrow 64 = 2 \times 2 \times 16$
$\Rightarrow 64 = 2 \times 2 \times 4 \times 4$
$\Rightarrow 64 = 4 \times 4 \times 4$
We have the exponential rule which is given by $\underbrace {a \times a \times a \times ....}_{n - times} = {a^n}$ ……(1)
Hence we can write 64 as, $64 = {4^3}$.
Now we consider the constant term which is 27.
We know that 27 can be written as,
$27 = 3 \times 3 \times 3$
Hence by exponential rule given by the equation (1), we have,
$27 = {3^3}$.
We have the exponential rule ${a^k}{b^k} = {\left( {ab} \right)^k}$, hence we get,
$64{x^3} - 27 = {\left( {4x} \right)^3} - {3^3}$
Now we apply the formula, ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$.
Here $a = 4x$ and $b = 3$
Hence we get,
$64{x^3} - 27 = \left( {4x - 3} \right)\left( {{{\left( {4x} \right)}^2} + 4x \times 3 + {3^2}} \right)$
$\Rightarrow 64{x^3} - 27 = \left( {4x - 3} \right)\left( {16{x^2} + 12x + 9} \right)$
Therefore the factorization of $64{x^3} - 27$ is given by $\left( {4x - 3} \right)\left( {16{x^2} + 12x + 9} \right)$.

Note:
It is important to remember the algebraic formulas of addition and difference of cubes. They are given by
${a^3} + {b^3} = (a + b)({a^2} - ab - {b^2})$
${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
Also note that when we multiply the above obtained factors, we need to get the given equation as result. If we don’t get the given equation as a result, then our solution is not correct. Also we must know the values of cubes of numbers as it simplifies our work.