CUET Preparation Today
If $(k + \frac{1}{k})^2 $ = 9, then what is the value of $k^3 +\frac{1}{k^3}$ ? |
27 18 15 21 |
18 |
If x + \(\frac{1}{x}\) = n then, $x^3 +\frac{1}{x^3}$ = n3 - 3 × n If $(k + \frac{1}{k})^2 $ = 9 $(k + \frac{1}{k}) $ = 3 Then, $k^3 +\frac{1}{k^3}$ = 33 - 3 × 3 = 18 |
