Answer:
123
Step-by-step explanation:
Squaring the given equation gives
[tex]t^2+(2\cdot t \cdot \frac{1}{t} )+ \frac{1}{t^2}=9.[/tex]
The [tex]t[/tex] and [tex]\frac{1}{t}[/tex] terms cancel out nicely, which is one of the reasons for squaring the equation.
Simplifying gives
[tex]t^2+2+\frac{1}{t^2}=9,[/tex]
[tex]t^2+\frac{1}{t^2}=7.[/tex]
The question asks for [tex]t^5+\frac{1}{t^5},[/tex] so we can square the equation again and simplify to get higher powers into the expression:
[tex](t^2+\frac{1}{t^2})^2=49,[/tex]
[tex]t^4+(2 \cdot t^2 \cdot \frac{1}{t^2}) + \frac{1}{t^4}=49,[/tex]
[tex]t^4+2+ \frac{1}{t^4}=49,[/tex]
[tex]t^4+\frac{1}{t^4}=47.[/tex]
Multiplying this expression by [tex]t+\frac{1}{t}[/tex] to try and get a fifth power gives
[tex](t^4+\frac{1}{t^4})(t+\frac{1}{t})=t^5+t^3+\frac{1}{t^3}+\frac{1}{t^5}.[/tex]
The only thing left we need is [tex]t^3+\frac{1}{t^3}[/tex] to subtract from this; we know everything else. Since [tex]t^3+\frac{1}{t^3}[/tex] can be written as [tex](t+\frac{1}{t})(t^2-1+\frac{1}{t^2}),[/tex] we can simply plug in the values we know for [tex]t+\frac{1}{t}[/tex] and [tex]t^2+\frac{1}{t^2}:[/tex]
[tex]t^3+\frac{1}{t^3}=(3)(7-1)=18.[/tex]
All that is left is to plug it in our equation here:
[tex](t^4+\frac{1}{t^4})(t+\frac{1}{t})=t^5+t^3+\frac{1}{t^3}+\frac{1}{t^5},[/tex]
[tex](47)(3)=t^5+\frac{1}{t^5}+18,[/tex]
Multiplying and rearranging gives:
[tex]t^5+\frac{1}{t^5}=123.[/tex]
