mau5 wrote:

Qoofi wrote:

\(a^4 + 1/ (a^4)\)= 119 and a > 1, then the value of \(a^3 - 1/(a^3)\)is :

A. 27

B. 36

C. 45

D. 54

E. 63

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.

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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