SOLUTION
The given question is
[tex]27x^3-64y^3[/tex]Simplify the expression
[tex]3^3x^3-4^3y^3=(3x)^3-(4y)^3[/tex]The difference of two cubes is given as
[tex]x^3-y^3=\mleft(x-y\mright)\mleft(x^2^{}+xy+y^2\mright)[/tex]Hence the given expression becomes
[tex](3x)^3-(4y)^3=(3x-4y)((3x)^2+3x(4y)+(4y)^2)[/tex]This gives
[tex](3x-4y)(9x^2+12xy+16y)[/tex]