Question

How do you factor: \[27{x^3} - 8{y^3}\]?

Answer 1

Hint: Here we will convert into the cubic form in the given terms of given expression. On doing some simplification we get the required answer.

Formula used: As the given terms are in cubic form, we will use the following formula:
\[a{}^3 - {b^3} = (a - b)({a^2} + ab + {b^2}).\]
If it is possible to do the middle term factorisation for \[({a^2} + ab + {b^2})\], then we will further perform it also.

Complete step-by-step solution:
The given expression is \[27{x^3} - 8{y^3}\].
Here, we can write \[27{x^3}\] as \[({3^3} \times {x^3})\].
So, it will become a form of cubic term.
It can be rewritten as \[{(3x)^3}\].
Similarly, \[8{y^3}\]can be writing as\[({2^3} \times {y^3})\]
So, it will also become in cubic form, i.e. \[8{y^3} = {(2y)^3}\].
So, we can rewrite the given expression as following:
\[ \Rightarrow 27{x^3} - 8{y^3}\]
\[ \Rightarrow {(3x)^3} - {(2y)^3}.\]
Now, it is clearly visible that we can put the above cubic formula to factorise these terms into two terms.
So, after using the formula, we get:
\[ \Rightarrow 27{x^3} - 8{y^3}\]
\[ \Rightarrow {(3x)^3} - {(2y)^3}\]
Now, using the cubic formula, we get:
\[ \Rightarrow 27{x^3} - 8{y^3}\]
\[ \Rightarrow {(3x)^3} - {(2y)^3}\]
\[ \Rightarrow (3x - 2y)\{ {(3x)^2} + (3x \times 2y) + {(2y)^2}\} .\]
Though, the \[(2{x^2} - 3{y^2})\] is in linear form, so we will leave this term as it is.
Now, after perform the multiplication and squared terms, we get the following expression:
\[ \Rightarrow 27{x^3} - 8{y^3}\]
On rewriting the terms and we get,
\[ \Rightarrow {(3x)^3} - {(2y)^3}\]
Using the formula we get,
\[ \Rightarrow (3x - 2y)\{ {(3x)^2} + (3x \times 2y) + {(2y)^2}\} \]
On rewriting we get
\[ \Rightarrow (3x - 2y)(9{x^2} + 6xy + 4{y^2}).\]
But, it is clear that the term \[(9{x^2} + 6xy + 4{y^2})\] cannot factorise further as the middle term of the expression cannot split into the two terms of the L.C.M of \[9,4\].

\[\therefore \]The final factorization of \[27{x^3} - 8{y^3}\] is \[(3x - 2y)(9{x^2} + 6xy + 4{y^2}).\]

Note: Points to remember:
\[1.\] If we have any trinomial with one squared term, we need to convert this expression into a quadratic expression \[(a{x^2} + bx + c)\], before we perform any further simplification.
\[2.\] Middle term of a quadratic expression\[(a{x^2} + bx + c)\], \[bx\], can be written as the sum of two different or same terms those are the factors of \[(a \times c)\].