How do you factor the expression \[{y^3} - 27\]?

Hint: Write the given expression and factorise the second term, then factorize the expression by the use of algebraic identity that gives the factors for the difference of two perfect cubes.
\[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\]

Complete step by step solution:
Write the given expression and factorise the second term as shown below.
\[{y^3} - 3 \cdot 3 \cdot 3\]
\[ \Rightarrow {y^3} - {3^3}\]

Use the algebraic identity \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\] to factorise the expression as follows:

Take \[y\] as \[a\] and \[3\] as \[b\], then substitute in \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\] as shown below
\[{y^3} - {3^3} = \left( {y - 3} \right)\left( {{y^2} + y\left( 3 \right) + {3^2}} \right)\]

Simplify each term of the expression as shown below.
\[ \Rightarrow \left( {y - 3} \right)\left( {{y^2} + 3y + 9} \right)\]

Therefore, the factorisation of \[{y^3} - 27\] is \[\left( {y - 3} \right)\left( {{y^2} + 3y + 9} \right)\].

Additional information: The proof for the algebraic identity \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\] is shown below.

Take the right hand side of the identity and simplify by the use of polynomial multiplication as shown below.
\[RHS = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\]

Multiply each term of the quadratic equation with binomial \[\left( {a - b} \right)\] as shown below.
\[ \Rightarrow {a^2}\left( {a - b} \right) + ab\left( {a - b} \right) + {b^2}\left( {a - b} \right)\]

Now, simplify each term further by multiplication as shown below.
\[ \Rightarrow {a^2}a - {a^2}b + aba - abb + {b^2}a - {b^2}b\]

Recollect and rearrange the expression as shown below.
\[ \Rightarrow {a^{2 + 1}} - {a^2}b + {a^{1 + 1}}b - a{b^{1 + 1}} + a{b^2} - {b^{2 + 1}}\]
\[ \Rightarrow {a^3} - {a^2}b + {a^2}b - a{b^2} + a{b^2} - {b^3}\]

It is observed that second and third terms are the same but one is negative and one is positive respectively.
\[\therefore {a^3} - a{b^2} + a{b^2} - {b^3}\]

Again, the second and third terms are the same but one is negative and other is positive respectively.
\[\therefore {a^3} - {b^3}\]

Therefore, equation \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\] is true.

Similarly, the proof for the algebraic identity \[{a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\] is shown below.

Take the right hand side of the identity and simplify by the use of polynomial multiplication as shown below.
\[RHS = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\]

Multiply each term of the quadratic equation with binomial \[\left( {a + b} \right)\] as shown below.
\[ \Rightarrow {a^2}\left( {a + b} \right) - ab\left( {a + b} \right) + {b^2}\left( {a + b} \right)\]

Now, simplify each term further by multiplication as shown below.
\[ \Rightarrow {a^2}a + {a^2}b - aba - abb + {b^2}a + {b^2}b\]

Recollect and rearrange the expression as shown below.
\[ \Rightarrow {a^{2 + 1}} + {a^2}b - {a^{1 + 1}}b - a{b^{1 + 1}} + a{b^2} + {b^{2 + 1}}\]
\[ \Rightarrow {a^3} + {a^2}b - {a^2}b - a{b^2} + a{b^2} + {b^3}\]

It is observed that second and third terms are the same but one is negative and one is positive respectively.
\[\therefore {a^3} - a{b^2} + a{b^2} + {b^3}\]

Again, the second and third terms are the same but one is negative and other is positive respectively.
\[\therefore {a^3} + {b^3}\]

Therefore, equation \[{a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\] is true.

Note: To solve any equation, the first step is to simplify and write the equation in standard form and then use the appropriate algebraic identity to factorise.
The quadratic expressions \[{a^2} + ab + {b^2}\] and \[{a^2} - ab + {b^2}\] have no real roots and so, it cannot factorise further.