Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

plse solve this

OpenStudy (loser66):

let t, r are the roots of binomial, you have formula that says t+r = -a t*r = b and t= r^2 there fore \[r^2+r=-a\\r*r=r^2 =b\]

OpenStudy (loser66):

get a = (-r^2 -r) --> a^3 =... b^2 =.... b = ...above --------------------------------------------- a^3+b^2+b =.......... do the leftover. quite easy, boy!

OpenStudy (anonymous):

if u have taken t = r^2 then it would be r^2*r not r*r

OpenStudy (loser66):

yes, fix it , thank you

OpenStudy (anonymous):

hmmm yes i fix it , but not getting answer

OpenStudy (loser66):

ok. So, I'm useless. Sorry for that.

OpenStudy (anonymous):

oh no y u said that no, u r not useless bro

OpenStudy (anonymous):

may be i will be wrong

OpenStudy (loser66):

to me, If I had that problem, I would do on that way. It's your problem, not mine,so I didn't solve until the final answer.But if you said so. I will try once and let you know whether I am right or wrong. That's it. no answer. Deal?

OpenStudy (anonymous):

okay

OpenStudy (loser66):

I am terribly sorry, friend. I AM RIGHT. hehehe.

OpenStudy (loser66):

you have to redo and find out your mistake. One more offer: I can check your work.

OpenStudy (anonymous):

okay ..i m attaching my answer

OpenStudy (loser66):

mistake at p^3 , it's still have minus sign in the front,

OpenStudy (loser66):

you let \[p^3 = (\beta^2+\beta)^3\]while it is \[p^3 = -(\beta^2+\beta)^3\]

OpenStudy (loser66):

ok? redo, friend

OpenStudy (anonymous):

yes.. but at the end of it cube, i have taken minus sign common

OpenStudy (loser66):

nope, pleeeeeaaase!! carefully put it in neat you will get the same answer with the first part you did. Sure!!

OpenStudy (loser66):

see? redo!! hehehe

OpenStudy (anonymous):

i m not getting can u help me

OpenStudy (loser66):

\[\huge p^3+q^2+q=( -(\beta^3+\beta))^3+(\beta^3)^2+\beta^3\] \[\huge -(\beta^6+3\beta^5+3\beta^4+\beta^3)+\beta^6+\beta^3\] \[open~the~bracket~\huge -\beta^6-3\beta^5-3\beta^4-\beta^3+\beta^6+\beta^3\] \[\huge -3\beta^5-3\beta^4\] \[factor~-3\beta^4~out~\huge -3\beta^4(\beta+1)\]

OpenStudy (loser66):

It's what you get from 3pq above. Therefore, the proof done.

OpenStudy (anonymous):

now, i got it where did i did mistake @Loser66 thanks bro

OpenStudy (loser66):

I really don't want to say "your welcome" at all. You NEDD PRACTICE!!! Unbelievable me. I will avoid you at all costs to not ruin your math skill. By giving you anything like this. I ruin your life. Sorry.!! just this time

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!