Latest Elitmus Aptitude Question SOLUTION: If (a^4)+(1/a^4)=119 then find value of (a^3)-(1/a^3)=????

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If (a^4)+(1/a^4)=119 then find value of (a^3)-(1/a^3)=????

Asked In Elitmus Prashant Baghel (9 years ago)

Solved gautam gaurav Read Solution (11)

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Read Solution (11) :

gautam gaurav  (9 years ago)

(a^4)+1/(a^4)=(a*a+1/a*a)(^2)-2=119
after solving we get as
(a-1/a)=3
so the value of (a^3)-(1/a^3)=(3)(1+11)=36
so ans is 36

Is this solution Helpfull? Yes (49) | No (7)  

Nisar Ali Khan  (9 years ago)

(a^4 + 1/a^4) = 119
(a^2 + 1/a^2)^2 = 119 + 2 = 121 => a^2 + 1/a^2 = 11
another eqn,
(a - 1/a) = sqrt(9) = 3

now,
a^3 - 1/a^3
= (a - 1/a)*(a^2 + a*1/a + 1/a^2)
= 3*(11+1)
= 36

Is this solution Helpfull? Yes (27) | No   

Deb Das  (9 years ago)

a^4+1/a^4=119
(a^2+1/a^2)^2-2.a.1/a=119
a^2+1/a^2=11
(a-1/a)^2+2.a.1/a=11
a-1/a=3

Now we Know a3 - b3 = (a - b)3 + 3 a b (a - b)
so,
(a^3)-(1/a^3)=(a-1/a)^3-3*a*1/a(a-1/a)=3^3-3*3=27-9=18
So the answer is 18

Is this solution Helpfull? Yes (26) | No (22)  

yogesh  (9 years ago)

a-1/a=3;
a^2+1/a^2=11;
so,
a^3-1/a^3=36

Is this solution Helpfull? Yes (1) | No   

Narasapu Hari prasad  (9 years ago)

Aaproxmatly 36

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

rakesh kumar singh  (9 years ago)

36 is the correct answer

Is this solution Helpfull? Yes (1) | No   

SWAPNESH GUPTA  (9 years ago)

(a+b)^2= a^2+b^2+2ab
put b=1/a

(a+1/a)^2=a^2+1/a^2+2

(a+1/a)^2-2=a^2+1/a^2
use this formula u will get the answer=36
firstly +2 use u will get 121
& then -2 use u will get 9

Is this solution Helpfull? Yes (1) | No   

vivek shukla  (9 years ago)

CANNOT BE DETERMINED??

Is this solution Helpfull? Yes | No (3)  

SYED MUZZAMMIL NAQVI  (9 years ago)

Ans- 36 is correct..This question came in Bank exam SBI PO-2013.

Is this solution Helpfull? Yes | No   

Anuj Awasthi  (9 years ago)

(a^4)+1/(a^4)=(a*a+1/a*a)(^2)-2=119
after solving we get as
(a-1/a)=3
so the value of (a^3)-(1/a^3)=(3)(1+11)=36
so ans is 36
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samm  (8 years ago)

(x^2+1/x^2)^2 = x^4 + 1/x^4 + 2*x^2*1/x^2
= 119 +2
= 121
x^2+1/x^2 = 11
(x ­ 1/x)^2 = x^2 + 1/x^2 ­ 2*x*1/x
= 11 ­2
= 9
x ­ 1/x = 3
now (x­1/x)^3 = x^3 ­ 1/x^3 ­3*x*1/x(x ­1/x)
so x^3 ­ 1/x^3 = 3^3 + 3*3
= 27 +9
= 36
final ans is 36

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