If $x+\frac{1}{x}=a, x^2+\frac{1}{x^3}=b$, then $x^3+\frac{1}{x^2}$ is

  • [KVPY 2011]
  • A

    $a^3+a^2-3 a-2-b$

  • B

    $a^3-a^2-3 a+4-b$

  • C

    $a^3-a^2+3 a-6-b$

  • D

    $a^3+a^2+3 a-16-b$

Similar Questions

Let $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$ be the solution of the equation $4 x^4+8 x^3-17 x^2-12 x+9=0$ and $\left(4+x_1^2\right)\left(4+x_2^2\right)\left(4+x_3^2\right)\left(4+x_4^2\right)=\frac{125}{16} m$. Then the value of $\mathrm{m}$ is..........

  • [JEE MAIN 2024]

If the sum of the two roots of the equation $4{x^3} + 16{x^2} - 9x - 36 = 0$ is zero, then the roots are

If $\alpha ,\beta $are the roots of ${x^2} - ax + b = 0$ and if ${\alpha ^n} + {\beta ^n} = {V_n}$, then

Let $\alpha$ and $\beta$ be the roots of the equation $\mathrm{x}^{2}-\mathrm{x}-1=0 .$ If $\mathrm{p}_{\mathrm{k}}=(\alpha)^{\mathrm{k}}+(\beta)^{\mathrm{k}}, \mathrm{k} \geq 1,$ then which one of the following statements is not true?

  • [JEE MAIN 2020]

Let $\alpha, \beta$ be roots of $x^2+\sqrt{2} x-8=0$. If $\mathrm{U}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}$, then $\frac{\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9}{2 \mathrm{U}_8}$ is equal to ............

  • [JEE MAIN 2024]