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If x+(1/x)=a,x^2+(1/x^3)=b then x^3+(1/x^2) is?how got it?

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Ask this again on the Homework Help category. And do it in the daytime.
(USA daytime) Then there will be people awake.

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Again these expansions:(a+b)^3=a^3+b^3+3ab^2+3a^2b and (a+b)^2=a^2+b^2+2ab so now we start!!!!:(x+1/x)^2=x^2+1/x^2+2(2*x*1/x)so x^2+1/x^2is a^2-2 and according to this x^3+1/x^3is a^3-3a^2+6!!!so if we add b and the Q we are gonna have that x^3+1/x^2+1/x^3+x^2 equals a^3-2a^2+4 So the answer is a^3-2a^2+4-b

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what is ur real name?
ur job?????
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Saharnaz im a student 16
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where r u from?
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have u face book?
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I carried 1000Kg of water melon in summer by train.in beginning the water content was 99%.By the time i reached the destination the water content had dropped to 98%.The reduction in the weight of water melon was?
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Im iranian!!yeah i do but i dont use it so often!!my kik is saharnax!!where R u from.
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Idk this question sry!!
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I think 0,01*1000=10kg
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i am Indian,Kerala
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let S={1,2,3...40} and let A be the subset of S such that no 2 element in A have their sum divisible by 5.What is the maximum no.of elements possible in A?
OPTIONS 10,13 ,17,20
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To workers A and B are engaged to do a piece of work.Working alone,A take 8 hours more to complete the work than if both worked together.Working alone, B would need 4 1/2(4 and half) hours more to complete the work than if both worked together. How much time would they take to complete the job working together?
options:4 hours,5 hours,6 hours,7 hours.
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The answer to the second one is 6!!!az we know if A does the whole job in a hours in one hour 1/a of the job is done and forB its 1/b so both of them joined together in one hour do1/a+1/b job so the whol work done by both of them is done in 1/1/a+1/b!!!so a-8=1/(1/a+1/b)=b-4.5!!now its just solving equations a-b equals 3,5so b equals a-3,5soo 1/a+1/bequals 1/a+1/a-3,5equals 1/a-8!!!if u solve this ur gonna have a=2and a=14because b=a-3,5 a equals 2is not accepted So A=14 B=10,5and the whole work is done in 6hours
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The answer to the first one i think is 17!!because suppose we start from having 1in iour subset!!so numbers 4814....39which are 8numbers cannot be included!!for number 2these 8numbers star from 3 8....38!!number 3and 4are already excluded so now number 5:10 15 20 25 30 35!!are excluded(6numbers)from 5to end except these 8+8+6numbers all vcan be included because forexample for 6we already have omited 9 14...and etc for previous numbers and remeber number 40itself cannot be used So 40-(8+8+6)-1=17
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Sorry i just dont know how to explain math in English!!
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Did u get them?
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To workers A and B are..................
this i understand..
but set i don't understand
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:a^3-b^3=(a-b)(a^2+b^2+ab) and (a-b)^2=a^2+b^2-2ab so now we start:512^3-259^3=(512-259)(512^2+259^2+512*259)-(512-259=253)^3--253(512^2+259^2+512*259-(512-259)^2)=253*3*512+259=11*13*2^9*3*7*37 So it has 6 prime divisors
pls explain once more how got it???????????
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In triangle ABC the altitude from B and C to opposite side are not shorter than their opposite sides.Then one of the angle in ABC is30,45,60,72
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