If \(x^2-7x-1=0\) and \(x>1\), then the value of \(x^3-\displaystyle\frac{1}{x^3}\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 364
Dividing the equation by \(x\) gives \(x-\frac{1}{x}\), and the cube follows from \(x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)\).
Dividing \(x^2-7x-1=0\) by \(x\): \(x-7-\displaystyle\frac{1}{x}=0\), so \(x-\displaystyle\frac{1}{x}=7\)
\(x^3-\displaystyle\frac{1}{x^3}=7^3+3\times7\)
\(=343+21=364\)