Answer:
[tex]27y^{3} +125z^{3} =(3y)^{3} +(5z)^{3}[/tex]
Apply the formula for sum of cubes: [tex]a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})[/tex]
Therefore:
[tex]\left(3y\right)^3+\left(5z\right)^3=\left(3y+5z\right)\left(3^2y^2-3\cdot \:5yz+5^2z^2\right)[/tex]
[tex]=\left(3y+5z\right)\left(9y^2-15yz+25z^2\right)[/tex]
