Question

hey mate
factorise ....

Answer 1

Answer:

Step-by-step explanation:

Both of them are in the form of a³+b³ which can be written as (a+b)(a²-ab+b²)

                             Therefore in the 1st one a=3y and b=5z

                       Can be factorized as (3y+5z)(9y²-15yz+25z)

For the 2nd one  

                              a = 4m and b = -7n

             Using the above formula

                           ⇔(4m-7m)(16m²+28mn+49n²)

Answer 2

$( i ) \\ \\ 27y^3 + 125z^3 \\ \\ = (3y)^3 + (5z)^3 \\ \\ = (3y + 5z)((3y)^2 - (3y \times 5z) + (5z)^2) \\ \\ = (3y + 5z)(9y^2 - 15yz + 25z^2) \\ \\ [ The\ identity\ a^3 + b^3 = (a + b)(a^2 - ab + b^2)\ is\ used\ here. ]$

$\\ \\ \\ ( ii ) \\ \\ 64m^3 - 343n^3 \\ \\ = (4m)^3 - (7n)^3 \\ \\ = (4m - 7n)((4m)^2 + (4m \times 7n) + (7n)^2) \\ \\ = (4m - 7n)(16m^2 + 28mn + 49n^2) \\ \\ [ The\ identity\ a^3 - b^3 = (a - b)(a^2 + ab + b^2)\ is\ used\ here. ]$