Question

Prime Factors of 125-y3.

Answer 1

we have a Edit Question button you know :)

Answer 2

anyway..use the formula for difference of two cubes here

\[\large a^3 - b^3 = (a-b)(a^2 + ab + b^2)\]

Answer 3

\[(5^3 - y^3)\]

Use the above formula to solve it...

Answer 4

How about the prime factors of \[x ^{3}-1\] ?

Answer 5

Just same..

You will use the same formula for this..

\[(x^3 - 1^3) = ???\]

Use the above formula to find it..

Answer 6

what is the prime actors of 8-\[x ^{3}\] ?

Answer 7

still same.. 8 is 2^3

Answer 8

x^3 is x^3

Answer 9

Using same formula :
\[2^3 - x^3\]

Answer 10

aa .. ThankYou ! :]

Answer 11

Welcome dear..

Answer 12

how about the product o monomial by a binomial ? 3(x+5) ?

Answer 13

Just distribute 3 inside the brackets:

\[a(b+c) = ab + ac\]

Answer 14

is my answer right ? 3x+15 ?

Answer 15

Yes you are right...

Answer 16

yey :]] and how about 4x(x-3) = 4x\[^{2}\]-12x ? is that's right ?

Answer 17

Yes have faith on yourself you are very much correct in doing this..

Answer 18

heh good one @eiznehr.12 :D

Answer 19

hehe :]] Thankyou .. i just multiplied the monomial to the terms inside the parenthesis ? :]

Answer 20

Yes that is what you are supposed to do...

Answer 21

but I don't know how to factorize \[x ^{3}\]-8 ..

Answer 22

difference of two cubes still...x^3 is x^3 and 8 is 2^3

Answer 23

\[(x^3 - 8) = (x^3 - 2^3)\]

Here, a = x and b = 2

Put these values of a and b in the formula written above and try it once on your own..

Answer 24

hmmmm .. i got (x-2)(y2+2x+4)

Answer 25

How come y^2??

Answer 26

You almost did that right but one mistake only...

Answer 27

sorry it's x :]

Answer 28

hehe

Answer 29

x or x^2???

Answer 30

its x^2

Answer 31

Yes now you are right:

\[x^3 - 8 = x^3 - 2^3 = (x-2)(x^2 + 2x + 4)\]

Answer 32

yes ! :]] ThanksALot :]]

Answer 33

Welcome dear..

Answer 34

ThankYou again :] hehe

Answer 35

Welcome again...