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    If x=(2−3)−1,x = \left(2 - \sqrt{3}\right)^{-1}, x=(2−3​)−1,​ then the value of x3−2x2−7x+5x^3 - 2x^2 - 7x + 5 x3−2x2−7x+5​ is-
    Question

    If x=(23)1,x = \left(2 - \sqrt{3}\right)^{-1}, ​ then the value of x32x27x+5x^3 - 2x^2 - 7x + 5 ​ is-

    A.

    2

    B.

    1

    C.

    0

    D.

    3

    Correct option is D

    Given:
    x=(23)1=123x = \left(2 - \sqrt{3}\right)^{-1} = \frac{1}{2 - \sqrt{3}}​​
    Rationalizing the denominator,
    x=123×2+32+3=2+3x = \frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = 2 + \sqrt{3}​​

    Formula used:
    We will evaluate x32x27x+5x^3 - 2x^2 - 7x + 5 ​ by substitution.
    Solution:
    Let x=2+3x = 2 + \sqrt{3}​​
    x2=(2+3)2=4+43+3=7+43x3=(2+3)(7+43)=14+83+73+12=26+153x^2 = (2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \\x^3 = (2 + \sqrt{3})(7 + 4\sqrt{3}) = 14 + 8\sqrt{3} + 7\sqrt{3} + 12 = 26 + 15\sqrt{3}​​
    Now substitute:
    x32x27x+5=(26+153)2(7+43)7(2+3)+5=26+15314831473+5=(261414+5)+(1538373)=3+03=3x^3 - 2x^2 - 7x + 5 \\= (26 + 15\sqrt{3}) - 2(7 + 4\sqrt{3}) - 7(2 + \sqrt{3}) + 5 \\= 26 + 15\sqrt{3} - 14 - 8\sqrt{3} - 14 - 7\sqrt{3} + 5 \\= (26 - 14 - 14 + 5) + (15\sqrt{3} - 8\sqrt{3} - 7\sqrt{3}) \\= 3 + 0\sqrt{3} = 3​​
    Correct answer is (d) 3

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