testing

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

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