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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55% (02:37) correct 
