Question

factorise27y^3+125z^3​

Answer 1

Question:

Factorise the given equation

Answer:

$(3y+5z)(9y^2)-5yz+(25z^2)$

Explanation:

The given equation is

$27 {y}^{2} + 125 {z}^{2}$

Hence the numbers has perfect cube  numbers

$\sqrt[3]{27} = 3$

$\sqrt[3]{125} = 5$

Now you can change them into while cube

${3y}^{3} + {5z}^{3}$

We can see that it is there in the formula that is

${a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )$

So using the formula we will put apply to the equation

${3y}^{3} + {5z}^{3} =(3y+5z)[(3y)^2-(3y)(5z)+(5z)^2$

⇒$(3y+5z)[(3y)^2-(3y)(5z)+(5z)^2$

⇒$(3y+5z)(9y^2)-5yz+(25z^2)$

Hence your answer is $(3y+5z)(9y^2)-5yz+(25z^2)$.