3) 64j^15 + 125y^21
Notice that this is just the sum of two cubes....and we have
Remember that
a ^3 + b^3 = ( a + b) (a^2 - ab + b^2)
So...let a = 4j^5 b = 5y^7 a^2 = 16j^10 b ^2 = 25y^14 and -ab = - 4j^5*5y^7
And we have.....
( 4j^5 + 5y^7) ( 16j^10 - 4j^5*5y^7 + 25y^14) =
(4j^5 + 5y^7) ( 16j^10 - 20 j^5y^7 + 25y^14)
4) c^3d^3 + 125 = (cd)^3 + 125
This is similar......
a ^3 + b^3 = ( a + b) (a^2 - ab + b^2)
a = cd b = 5 a^2 = (cd)^2 - ab = - 5cd and b^2 = 25
So we have
( cd + 5) [ (cd)^2 - 5cd + 25]