Latest Elitmus Aptitude Question SOLUTION: If a^4 - 1/a^4 =119 what is the value of a^3 – 1/ a^3 ?

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If a^4 - 1/a^4 =119 what is the value of a^3 – 1/ a^3 ?

Asked In Elitmus Narendra Pargain (10 years ago)

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Sahil  (10 years ago)

question had ( a^4 + 1/a^4 )= 119.
nw...
(a^2+1/a^2)^2 -2=119
so (a^2+1/a^2)^2 = 121
so a^2+1/a^2=11

nw (a-1/a)^2 + 2 =11
(a-1/a)^2 =9
a-1/a=3
by cubing both sides (a-1/a)^3 = a^3 - 1/a^3 -3(a-1/a)
so 27= a^3 - 1/a^3 -9


a^3 - 1/a^3= 36 answer.... (:

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RAHUL BANSAL  (10 years ago)

first of all question had ( a^4 + a^4 )= 119.
now coming to solution.
(a^2 + 1/a^2)^2 = a^4 + 1/a^4 +2
solving the above equation gives, (a^2 + 1/a^2) = 11
now writing , (a - 1/a)^2 = a^2 + 1/a^2 -2
solving this equation gives , (a-1/a)=3
now writing, (a - 1/a)^3 = ( a^3 - 1/a^3) - 3(a - 1/a)
so (a^3 - 1/a^3) = 27+9 = 36
solving the above equation gives , (a^3 - 1/a^3) = 36
cheers!!!!!!

Is this solution Helpfull? Yes (4) | No (4)  

Narendra Pargain  (10 years ago)

If a^4 - 1/a^4 =119 what is the value of a^3 – 1/ a^3 ?

Is this solution Helpfull? Yes (3) | No (3)  

vaibhav sharma  (10 years ago)

first of all question had ( a^4 + a^4 )= 119.
now coming to solution.
(a^2 + 1/a^2)^2 = a^4 + 1/a^4 +2
solving the above equation gives, (a^2 + 1/a^2) = 11 this is because A.M>=G.M(i.e you took +ve value 11)
now writing , (a - 1/a)^2 = a^2 + 1/a^2 -2 =11-2=9
solving this equation gives , (a-1/a)=3 or -3 because it does not satisfy A.M>=G.M hence we cannot determine expression a^3-(1/a)^3 because for a-1/a=3 we get a^3-(1/a)^3=36 and for a-1/a=-3 we get a3-(1/a)^3=-36 and it was not given in the question that a>0 or real number or whole num something like that

Is this solution Helpfull? Yes (3) | No (1)  

VIKAS SINGH  (10 years ago)

since a^4-1/a^4=119 so (a^2-1/a^2).(a^2+1/a^2)=1*119 or 7*17.but when we take the value as 7*17 it does not setisfied so (a^2-1/a^2)=1 nd (a^2+1/a^2)=119 so a+1/a=11.
now from a^2-1/a^2=1 so a-1/a=1/11.
since a^3-1/a^3=(a-1/a)(a^2+1/a^2+1)=120/11.
so ans will be 120/11.

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RAHUL BANSAL  (10 years ago)

was there any mention of x being > 0

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