MCQ
If $\text{x}+\frac{1}{\text{x}}=3,$ then $\text{x}^6+\frac{1}{\text{x}^6}=$
  • A
    927
  • B
    414
  • C
    364
  • D
    322

Answer

  1. 322
    Solution:
    $\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
    $\text{x}+\frac{1}{\text{x}}=3$ (given)
    $\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=(3)^2-2$
    $\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=7\ ...(1)$
    Cubing both side of equation (1). we have
    $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^3=(7)^3$
    $\Rightarrow(\text{x}^2)^3+\Big(\frac{1}{\text{x}^2}\Big)^3+3(\text{x}^2)\frac{1}{\text{x}^2}\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=7^3$
    $\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}+3(7)=7^3$
    $\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=343-21$
    $\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=322$
    Hence, correct option is (d).

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