Please use the Math function to wrap the mathematical expressions.

Given that \(a^4 + \frac{1} {(a^4)}\) = 119 , adding 2 on both sides, we get : \([a^2 + \frac{1}{a^2}]^2\) \(= 121 \to a^2 + \frac{1}{a^2} = 11\)

Again, by subtracting 2 on both sides, we have \((a-\frac{1}{a})^2 = 9 \to (a-\frac{1}{a}) = 3\)

Now, \(a^3 - \frac{1}{a^3} = (a-\frac{1}{a})(a^2+\frac{1}{a^2}+1) = 12*3 = 36\)

B.
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Consider f(a) = a^4+(1/a)^4 = 119 and a>1

if a = 3, f(a) ~= 81
if a = 4, f(a) ~= 256

By this we can definitely say 3<a<4 and the value must definitely be closer to 3 (considering f(a) is closer to f(3)).

Substituting the value in
g(a) = a^3-(1/a)^3 ,
we could defintely say g(a) is slightly greater than g(3).
And, the most likely answer could be .

/SW

If you feel that algebra is too tedious, you can approximate the value of 'a' to get to the right answer. But you must find both the minimum and maximum value to be sure of the right answer.

You know that a > 1.
\(3^4 = 81\)
\(4^4 = 256\)

To get 119, a would be more than 3 but less than 4. It will be closer to 3. Say, it will be something like 3.2 or 3.3 etc.

\(3.5^2 = 12.25\) (Note that you can easily find the square of numbers ending in 5. You write 25 at the end and multiply the first digit with the next integer and write in front so \(35^2 = .....25\) and then \(3*4 = 12\) so 1225)

But 12*12 = 144 so a is obviously less than 3.5.

\(a^3\) will be more than \(3^3 = 27\) but less than \(3.5^3\) which is approximately \(12*3.5 = 42\).

Only option 36 lies in between these values.

Answer (B)

Note that 1/a^4 and 1/a^3 have no role to play in this method because they are very small (just a few decimal points) compared to the rest of the numbers.
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a^3 - 1/a^3 = (a - 1/a)(a^2 + 1/a^2 + 1)

So, we have to find the values of (a - 1/a) and (a^2 + 1/a^2)

a^4 + 1/a^4 = 119 will help us find this information.

a^4 + 1/a^4 = (a^2 + 1/a^2)^2 - 2

So, (a^2 + 1/a^2)^2 - 2 = 119
-> (a^2 + 1/a^2)^2 = 121
-> (a^2 + 1/a^2) = 11

Again, (a - 1/a)^2 = a^2 + 1/a^2 -2
-> (a - 1/a)^2 = 11 -2
-> (a - 1/a)^2 = 9
-> (a - 1/a) = 3

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

spent more than 40 minutes to figure out how to solve it..
unsuccessfully...
mau5 says that he added 2 on both sides, but doesn't show that he added on the left side ...
pretty confusing one..
can anyone explain through the algebra way???
Bunuel?

Here are more details to mau5's solution.

We need to find \(a^3 - \frac{1}{a^3}\)

\(a^3 - \frac{1}{a^3} = (a-\frac{1}{a})(a^2+\frac{1}{a^2}+1)\)

So we need the values of \(a-\frac{1}{a}\) and \((a^2+\frac{1}{a^2})\)

Given:
\(a^4 + \frac{1} {(a^4)} = 119\)

Add 2 on both sides to get:

\(a^4 + \frac{1} {(a^4)} + 2 = 119 + 2\)

which is equivalent to:
\(a^4 + \frac{1} {(a^4)} + 2*a^2*\frac{1}{a^2} = 121\)

Use \(x^2 + y^2 + 2xy = (x + y)^2\) to get

\([a^2 + \frac{1}{a^2}]^2 = 11^2\)

So, \([a^2 + \frac{1}{a^2}] = 11\)

Now, you have the value of one expression but one is still required.

Subtract 2 from both sides:

\([a^2 + \frac{1}{a^2} - 2] = 11 - 2\)

which is equivalent to:
\([a^2 + \frac{1}{a^2} - 2*a*\frac{1}{a}] = 9\)

Again, use \(x^2 + y^2 + 2xy = (x + y)^2\) to get

\((a-\frac{1}{a})^2 = 3^2\)

\((a-\frac{1}{a}) = 3\)

Now we have both the required quantities.

\(a^3 - \frac{1}{a^3} = (a-\frac{1}{a})(a^2+\frac{1}{a^2}+1) = 3*(11 + 1) = 36\)
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My solution is not as smart as the one above. But I suppose to use an approximation. if a>1 and a^4+1/(a^4)=119, so we know that 1/(a^4) is <1, so a^4~119, a^2~11.
Again, 1/a^3 is <1, so we need to calculate a^3=11*(11^0,5). What is the 11^0,5? - it is more than 3, but below 4. therefore the answer will be between 33 and 44, closer to 33. Thus, we have 36 (B)

why a-1/a is not equal to = -3 ??? please explain.

Value of a cannot be -3 because a should be greater than 1(a>1) as mentioned in question

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