How do you factor $64{x^3} - 125$?

Hint: In the question, we have an equation that is quadratic, and we have to find its factors, which can be done with the help of rules. Factorisation is the reverse of multiplying out. One important factorisation process is the reverse of multiplications. The factors of any equation can be an integer, a variable or an algebraic expression itself.

Formula used: ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$

Complete step-by-step solution:
We have an equation $64{x^3} - 125$, and we are supposed to factorise it,
Upon keen observation, we can conclude that this can be simplified into the formula that we have which goes like ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
We have $64{x^3}$ which can be written as ${(4x)^3}$ and we also have $125$ which can be written as ${(5)^3}$.
Putting the values in the formula we get,
$ \Rightarrow {(4x)^3} - {(5)^3}$We break it down and factorise it into $(4x - 5)({(4x)^2} + (4x \times 5) + {(5)^2})$
Opening the inner brackets to make it look a bit simplified, we get,
$ \Rightarrow (4x - 5)(16{x^2} + 20x + 25)$

Therefore, when the equation $64{x^3} - 125$ is simplified, we get its factors that are $ (4x - 5)(16{x^2} + 20x + 25)$

Note: In Mathematics, factorisation or factoring is defined as the breaking or decomposition of an entity (for example a number, a matrix, or a polynomial) into a product of another entity, or factors, which when multiplied together gives the original number or a matrix, etc. It is simply the resolution of an integer or polynomial into factors such that when multiplied together they will result in an original or initial integer or polynomial. In the factorisation method, we reduce any algebraic or quadratic equation into its simpler form, where the equations are represented as the product of factors instead of expanding the brackets.