Question

\[\frac{1}{a^3}+\frac{1}{b^3}=?\] if a+b= x and ab=y

Answer 1

help me please

Answer 2

take LCM, its done

Answer 3

\[\frac{a^3+b^3}{a^3b^3}\]

Answer 4

af

Answer 5

after that ?

Answer 6

wait one mint

Answer 7

x(b^2 -y +a^2)/a^3b^3

Answer 8

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

Answer 9

a^2b^2=(ab)^2

Answer 10

and put the given values

Answer 11

am not able to solve further please show the full process

Answer 12

okay

Answer 13

[a^2+b^2]/[a^2b^2]
=[a^2+b^2+2ab-2ab]/[(ab)^2]
=[(a+b)^2-2ab]/[(ab)^2]
=[x^2-2y]/[y]^2

Answer 14

done?

Answer 15

that is okay but it should be (ab)^3 or y^3 in the denominator but i understood the process thanks

Answer 16

u simplify, whatever is it

Answer 17

no no sorry i think the whole process went wrong zakaullah ,,,, i would request u to correct it and post it here please so that i can understand it !!!!!!!!!!!!!!!!

Answer 18

its correct, just think about it a little bit

Answer 19

\[\frac{a^3+b^3}{a^3b^3}\]
\[\frac{(a+b)(a^2+b^2+ab)}{a^3b^3}\]
now ?

Answer 20

here it looks like a^2, sorry for that

Answer 21

it's \(\frac{1}{a^3} + \frac{1}{b^3}\) right?

Answer 22

k no problem
is it right now :

\[\frac{a^3+b^3}{a^3b^3}\]
\[\frac{(a+b)(a^2+b^2+ab)}{a^3b^3}\]
\[\frac{(a+b){(a+b)^2-ab)}}{y^3}\]
\[\frac{x(x^2-ab)}{y^3}\]
\[\frac{x^3-xy}{y^3}\]

Answer 23

in the above
[a^2+b^2+ab]
=[a^2+b^2+2ab-ab]
=(a+b)^2-ab
now try

Answer 24

m i right ?

Answer 25

yes 100%

Answer 26

it is showing the answer as \[\frac{x^3-3xy}{y^3}\]

Answer 27

you didn't factor the sum of cubes correctly

Answer 28

why zarkon sir ?

Answer 29

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

Answer 30

k no problem
is it right now :

\[\frac{a^3+b^3}{a^3b^3}\]
\[\frac{(a+b)(a^2+b^2-ab)}{a^3b^3}\]
\[\frac{(a+b){(a+b)^2+3ab)}}{y^3}\]
\[\frac{x(x^2+3ab)}{y^3}\]
\[\frac{x^3-3xy}{y^3}\]

Answer 31

now it is right ?

Answer 32

the work is not totally correct

Answer 33

yes it may be some where wrong since i didn't correct it perfectly
ok zarkon sir thanks very much for your help ... greatful to you sir !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 34

hav to go sir bye

Answer 35

\[\frac{(a+b)(a^2+b^2-ab)}{a^3b^3}=\frac{(a+b)((a+b)^2-3ab)}{a^3b^3}=\cdots\]

Answer 36

yes sir i got it i put + and in answer i put - . thanks again

Answer 37

good