Problem
ALG-B1-M06-P015 Cube of a reciprocal sum
#15
★★★☆☆ Level 3 of 5
Let \(x\neq 0\) and \(x+\frac{1}{x}=3\). Find \(x^3+\frac{1}{x^3}\).
Cube \(x+\frac{1}{x}\).
\(\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)\). Hence \(27=x^3+\frac{1}{x^3}+9\), so the desired value is \(18\).
Repeats the example as an independent training problem.