If $\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{6}$, then the value of $x^6 +\frac{1}{x^6}$ will be : |
2270 2502 2702 2712 |
2702 |
If $\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{6}$, then the value of $x^6 +\frac{1}{x^6}$ = ? we know that, $(x - \frac{1}{x}) = a$ (x2 + x-2) = (a)2 + 2 = b If x + \(\frac{1}{x}\) = n then, $x^3 +\frac{1}{x^3}$ = n3 - 3 × n If $K+\frac{1}{K}=n$ then, $K^2+\frac{1}{K^2}$ = n2 – 2 According to the question, $\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{6}$ then, $x+\frac{1}{x}$ = $\sqrt{6}$2 – 2 = 4 and then, $x^3+\frac{1}{x^3}$ = 43 – 3 × 4 = 52 Now apply the formula of square again, $x^6 +\frac{1}{x^6}$ = 522 – 2 = 2702 |