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    Q. If a-\frac{1}{a}=7 then a^3-\frac{1}{a^3}= ?
    Solution: We are given,


        \[a-\frac{1}{a}=7\]

    [latex]
    cubing both sides, we get:

        \[\begin{aligned}& a^3-\frac{1}{a^3}-3 \cdot a \cdot \frac{1}{a}\left(a-\frac{1}{a}\right)=7^3=343 \& \Rightarrow a^3-\frac{1}{a^3}=343+3\left(a-\frac{1}{a}\right) \&=343+3 \cdot 7 \&=343+21 \&=364 .\end{aligned}\]


    Q. If a-\frac{1}{a}=6 then a^4+\frac{1}{a^4}= ?
    Solution

        \[a-\frac{1}{a}=6\]


    Squaring both sides, wa get

        \[\begin{aligned}& a^2+\frac{1}{a^2}-2 \cdot a \cdot \frac{1}{a}=6^2=36 \\Rightarrow & a^2+\frac{1}{a^2}=36+2=38\end{aligned}\]

    Again Squaring both sides, we get

        \[\begin{aligned}& a^4+\frac{1}{a^4}+2 \cdot a^2 \cdot \frac{1}{a^2}=(38)^2 \& \Rightarrow a^4+\frac{1}{a^4}=1444-2 \&=1442 .\end{aligned}\]


    Q. If a-\frac{1}{a}=7 then find

        \[\begin{aligned}& a^2+\frac{1}{a^2}\end{aligned}\]


    Solution: \quad a-\frac{1}{a}=7
    Squaring both Sides, we got

        \[\begin{aligned}& a^2+\frac{1}{a^2}-2 \cdot a \cdot \frac{1}{a}=7^2=49 \\Rightarrow & a^2+\frac{1}{a^2}=49+2=51 .\end{aligned}\]


    Q. If a-\frac{1}{a}=1, then find

        \[\begin{aligned}& a^2+\frac{1}{a^2}\end{aligned}\]


    Solution: squaring both sides, we get

        \[a^2+\frac{1}{a^2}-2 \cdot a \cdot \frac{1}{x}=1^2=1\]


        \[\Rightarrow a^2+\frac{1}{a^2}=1+2=3 \text { Ans. }\]


    Q. If x-\frac{1}{x}=3 then find x^4+\frac{1}{x^4}.
    Solution: We are given,

        \[x-\frac{1}{x}=3\]


    Squaring both bides, we get

        \[\begin{aligned}& \left(x-\frac{1}{x}\right)^2=3^2=9 \\Rightarrow & x^2+\frac{1}{x^2}-2 \cdot x \cdot \frac{1}{x}=9 \\Rightarrow & x^2+\frac{1}{x^2}=11\end{aligned}\]


    Again squaring both sides, we get

        \[\begin{aligned}& x^4+\frac{1}{x^4}+2 \cdot x^2 \cdot \frac{1}{x^2}=11^2=121 \& \Rightarrow x^4+\frac{1}{x^4}=121-2=119\end{aligned}\]

    Q. If x-\frac{1}{x}=3, find x^5-\frac{1}{x^5}.
    Solution:

        \[x^5-\frac{1}{x^5}=\left(x^2+\frac{1}{x^2}\right)\left(x^3-\frac{1}{x^3}\right)-\left(x -\frac{1}{x}\right)\]


    \because we are given,

        \[\begin{aligned}x-\frac{1}{x} & =3 \x^2+\frac{1}{x^2} & =3^2+2=11 \x^3-\frac{1}{x^3}-3 \cdot x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right)=3^3=27 & \\Rightarrow x^3-\frac{1}{x^3} & =27+3\left(x-\frac{1}{x}\right) \& =27+3.3 \& =27+9 \& =36\end{aligned}\]


    \therefore Plugging the values, we get
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    Q. \quad x-\frac{1}{x}=2. Find x^6+\frac{1}{x^6}
    Sol wo r \quad x-\frac{1}{x}=2
    squaring both sides, we got

        \[\begin{aligned}& x^2+\frac{1}{x^2}-2 \cdot x \cdot \frac{1}{x}=4 \\Rightarrow & x^2+\frac{1}{x^2}=4+2=6\end{aligned}\]


    Cubing both sides, we get

        \[\begin{aligned}& x^6+\frac{1}{x^6}+3 \cdot x^2 \cdot \frac{1}{x^2}\left(x^2+\frac{1}{x^2}\right)=6^3=216 \\Rightarrow & x^6+\frac{1}{x^6}+3 \cdot 6=216 \\Rightarrow \quad & x^6+\frac{1}{x^6}=216-18=\frac{198}{\text { Ans }} \& \text { other formats of } x-\frac{1}{x} \text { type }\end{aligned}\]


    # Other formats of x-\frac{1}{x} type
    Q. If x^2-3 x-1=0
    then the value of \left(x^2+8 x-1\right) \cdot\left(x^3+x^{-1}\right)^{-1}
    Solution: We are given a quadratic equation

        \[x^2-3 x-1=0\]


    If we divide whose equation thoughts, then

        \[\begin{aligned}& \frac{x^2-3 x-1}{x}=0 \\Rightarrow & x-3-\frac{1}{x}=0 \\Rightarrow & x-\frac{1}{x}=3\end{aligned}\]


    The expression that we have to find out is

        \[\frac{x^2+8 x-1}{x^3+\frac{1}{x}}\]


    If w divide both numerator and Denominator by x, then

        \[\Rightarrow \frac{\frac{x^2+8 x-1}{x}}{\frac{x^3+\frac{1}{x}}{x}}=\frac{\left(x-\frac{1}{x}\right)+8}{x^2+\frac{1}{x^2}}\]


    We have to first find out the value of x^2+\frac{1}{x^2}.

        \[x-\frac{1}{x}=3\]


    squaring both sides, we get

        \[\begin{aligned}& 1 x^2+\frac{1}{x^2}-2 \cdot x \cdot \frac{1}{x}=3^2=9 \& \Rightarrow \quad x^2+\frac{1}{x^2}=9+2=11\end{aligned}\]


    If we plug the value in expression then,

        \[\frac{3+8}{11}=\frac{11}{11}=1\]


    Q. If x\left(3-\frac{2}{x}\right)=\frac{3}{x} then x^2+\frac{1}{x^2}= ?
    Solverion: We have

        \[\begin{aligned}& x\left(3-\frac{2}{x}\right)=\frac{3}{x} \\Rightarrow & 3 x-\frac{2}{x} \cdot x=\frac{3}{x} \\Rightarrow & 3 x-\frac{3}{x}=2 \\Rightarrow & x-\frac{1}{x}=\frac{2}{3}\end{aligned}\]


    Squaring both sides, we get

        \[\begin{aligned}& x^2+\frac{1}{x^2}-2 \cdot x \cdot \frac{1}{x}=\left(\frac{2}{3}\right)^2=\frac{4}{9} \\Rightarrow & x^2+\frac{1}{x^2}=\frac{4}{9}+2=\frac{22}{9} \text { or } 2 \frac{4}{9} .\end{aligned}\]

    Q. If x\left(5-\frac{2}{x}\right)=\frac{5}{x}. find x^2+\frac{1}{x^2}.
    Ans. \frac{54}{25}

    Q. If x-\frac{1}{x}=5 then find

        \[\frac{x^6+3 x^3-1}{x^6-8 x^3-1}\]


    Solution: If we divide both numerator and denominator by x^3, then

        \[\frac{\left(x^3-\frac{1}{x^3}\right)+3}{\left(x^3-\frac{1}{x^3}\right)-8}\]


    We are given,

        \[x-\frac{1}{x}=5\]


    Cubing both sides, we get

        \[\begin{aligned}& x^3-\frac{1}{x^3}-3 \cdot x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right)=5^3=125 \& \Rightarrow \quad x^3-\frac{1}{x^3}=125+3\left(x-\frac{1}{x}\right) \&=125+3 \cdot 5 \&=125+15 \&=140\end{aligned}\]

        \[\frac{140+3}{140-8}=\frac{143}{132}=\frac{13}{12}\]


    Q. If x\left(3-\frac{2}{x}\right)=\frac{3}{x}. Find x^3-\frac{1}{x^3}.
    Ans. \quad \frac{62}{27}.

        \[x^5-\frac{1}{x^5}=\left(x^2+\frac{1}{x^2}\right)\left(x^3-\frac{1}{x^3}\right)\]


    How do we derive x^5-\frac{1}{x^5} ?

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