1. Find the HCF of the following numbers by prime factorisation and continued division method10,35,40​

Step-by-step explanation:

10 = 2*5

35 = 5*7

40 = 2*2*2*5

The number 5 is common to all three  factorisations

So the HCF = 5.

Continued division:

10 and 35:

 10) 35(3

       30

          5)10(2

              10

                0

- so the HCF of 10 and 35 is 5.

10 and 40:

10)40( 4

     40

        0  

So the HCF of 10 and 40 is 10

We found HCF of 10 and 35 is 5 so:

HCF of 10, 35 and 40 is HCF of 5 and 10 which is 5.  

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)