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Polynomials — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsPolynomials1 Mark
MCQ
If $\text{x}^3-\frac{1}{\text{x}^3}=14$ than $\text{x}-\frac{1}{\text{x}}=$
  • A
    4
  • B
    2
  • C
    3
  • D
    5
✓

Answer

  1. 2
    Solution:
    Given: $\text{x}^3-\Big(\frac{1}{\text{x}^3}\Big)=14$
    Let x = a and $\frac{1}{\text{x}}=\text{b}$
    Say, $\text{x}-\frac{1}{\text{x}}=\text{A}$
    Then, $\text{a}^3-\text{b}^3=14$
    $\Rightarrow(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)=14$
    $\Rightarrow(\text{a}-\text{b})(\{(\text{a}-\text{b})^2+2\text{ab}\}+2\text{ab})=14$
    $\Rightarrow(\text{a}-\text{b})\{(\text{a}-\text{b})^2+3\text{ab}\}=14$
    $\Rightarrow(\text{a}-\text{b})\{(\text{a}-\text{b})^2+3\}=14$
    $\Rightarrow\text{A}(\text{A}^2+3)=14$
    $\Rightarrow\text{A}(\text{A}^2+3)=14$
    $\Rightarrow\text{A}^3+3\text{A}-14=0$
    $\Rightarrow\text{A}^3-2\text{A}^2+2\text{A}^2-4\text{A}+7\text{A}-14=0$
    $\Rightarrow\text{A}^2(\text{A}-2)+2\text{Y}(\text{Y}-2)+7(\text{Y}-2)=0$
    $\Rightarrow(\text{A}-2)(\text{A}^2+2\text{A}+7)=0$
    $\Rightarrow\text{A}-2=0,$
    $\Rightarrow\text{A}=2$
    $\Rightarrow\text{x}-\frac{1}{\text{x}}=2$

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