Question

How do you factor: \[8{x^6} - 27{y^6}\]?

Answer 1

Hint: Here we have to factor the given expression. So, we do some formula for splitting the given expression as per we need to solve this. 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 \[8{x^6} - 27{y^6}\].
Here, we can write \[8{x^6}\] as \[{2^3} \times {({x^2})^3}\].
So, it will become a form of cubic term.
It can be rewritten as \[{(2{x^2})^3}\].
Similarly, \[27{y^6}\] can be write as \[{3^3} \times {({y^2})^3}\]
So, it will also become in cubic form, i.e. \[27{y^6} = {(3{y^2})^3}\].
So, we can rewrite the given expression as following:
\[8{x^6} - 27{y^6}\]
\[ \Rightarrow {(2{x^2})^3} - {(3{y^2})^3}\]
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:
\[8{x^6} - 27{y^6}\]
\[ \Rightarrow {(2{x^2})^3} - {(3{y^2})^3}\]
By using the formula on splitting this we get
\[ \Rightarrow (2{x^2} - 3{y^2})\{ {(2{x^2})^2} + (2{x^2} \times 3{y^2}) + {(3{y^2})^2}\} \]
Though, \[(2{x^2} - 3{y^2})\]is in squared form, but the constant terms cannot be squared, so we will leave this term as it is.
Now, after perform the multiplication and squared terms, we get the following expression:
\[ \Rightarrow (2{x^2} - 3{y^2})(4{x^4} + 6{x^2}{y^2} + 9{y^4})\]
So, it is not possible to write \[(2{x^2} - 3{y^2})\] into differences of two perfect square, but if we squared the term in following way, then there could be a another solution of it:
\[ \Rightarrow (2{x^2} - 3{y^2}) = {(\sqrt 2 x)^2} - {(\sqrt 3 y)^2}\].
So, it can be expressed in the formula of \[({a^2} - {b^2}) = (a - b)(a + b)\].
So, the above expression can be rewrite as:
\[(2{x^2} - 3{y^2}) = {(\sqrt 2 x)^2} - {(\sqrt 3 y)^2} = (\sqrt 2 x - \sqrt 3 y)(\sqrt 2 x + \sqrt 3 y)\].
So, the given expression can be further factorising as:
\[ \Rightarrow (\sqrt 2 x - \sqrt 3 y)(\sqrt 2 x + \sqrt 3 y)(4{x^4} + 6{x^2}{y^2} + 9{y^4})\]
But using the trial and error method the trinomial \[(4{x^4} + 6{x^2}{y^2} + 9{y^4})\] cannot factorise further into the multiplication of two terms.

\[\therefore \]\[8{x^6} - 27{y^6}\] can be factored into \[(\sqrt 2 x - \sqrt 3 y)(\sqrt 2 x + \sqrt 3 y)(4{x^4} + 6{x^2}{y^2} + 9{y^4}).\]

Note: Points to remember that:
\[1.\]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 \[(a \times c)\].