Question

Factorise completely 8p + 125py^3

Answer 1

Answer:

 8 p + 125py^3

= p ( 8 + 125y^3 )

= p [ 2^3 + ( 5y )^3 ]

= p [ (2+5y) (4 - 10y + 25y^2) ]

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

= p ( 5y + 2 ) ( 25y^2 - 10y + 4 )

hope it helps !!!!!