Latest Elitmus Aptitude Question SOLUTION: (a^4)+(1/a^4)=119 then (a^3)+(1/a^3)=? Don't remember the options.

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(#M40020985) ELITMUS question Keep an EYE

(a^4)+(1/a^4)=119
then (a^3)+(1/a^3)=?

Don't remember the options.

Asked In Elitmus Anmol Modi (10 years ago)

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siddanth  (10 years ago)

answer is 10*13^1/2
a^4 + 1/a^4 = (a^2 + 1/a^2)^2 -2(a^2)(1/a^2)
from above we get a^2 + 1/a^2 = 11
a^2 + 1/a^2 = (a + 1/a)^2 - 2(a)(1/a)
from this we get a+1/a= (13)^1/2

a^3 + 1/a^3 = (a +1/a)^3 -3(a)(1/a)[a + 1/a]
= 13*3^1/2 - 3*13^1/2
= 10*13^1/2

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sainath  (10 years ago)

The answer is 130.

Well there are two ways 2 solve this

The Regular calculation is,

(a^2)^2+(1/a^2)^2 =119 [since a^2+b^2 = (a+b)^2-2ab)]
(a^2+1/a^2)^2 - 2(a^2)(1/a^2) = 119
(a^2+1/a^2)^2 = 121
a^2+1/a^2=11 [since a^2+b^2 = (a+b)^2-2ab)] ----- (1)
(a+1/a)-2(a)(1/a) =11
(a+1/a) = 13 --------(2)

a^3+(1/a)^3 = (a+1/a)(a^2+(1/a)^2- a(1/a) )[since a^3+b^3 = (a+b)(a^2+b^2-ab)]
= (13)(11-1)[From (1) and (2) (a+1/a)=13 & a^2+1/a^2=11 ]
= (13)(10)
= 130


The short cut for the problem for any number in place of 119 is lets say n,
add 2 to the number and take square root of that number (sqrt(n+2))
and the answer is (sqrt(n+2)-1)(sqrt(n+2)+2) is the answer.

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