Question

If $x^2+\frac{1}{x^2}=7$ find the value of $x^3+\frac{1}{x^3}$.

Answer 1

$If{x}^2+1/x^2-7$ find the valve of $x^3+1/3^3$

Given equation is $x^2+1/x^2=7$

we know that$(x+1/x{)}^2=x^2+2.x.1/x+(1/x{)}^2$

$=x^2+2+1/x^2$

$=(x^2+1/x^2)+2$

$=7+2$

$=9.$

$\therefore (x+1/x{)}^2=9$

$\Rightarrow (x+1/x)={\sqrt{\left(3\right)}}^x$

$\Rightarrow (x+1/x)=3.$

Now ,

$(x^3+1/x{)}^3=(x+1/x{)}^3-3.x.1/x.(x+1/x)$

$=(x+1/x{)}^3-3(x+1/x)$

$=(3{)}^3-3(3)$

$=27-9$

$=18.$

$\therefore x^3+1/x^3=18.$