Grade 8 simple algebra equation help I find this question hard, please help.
It is given that $x+\frac{1}{x}=3$
and $x^2+\frac{1}{x^2}=7$.

Please find the value of $x^3+\frac{1}{x^3}$.
Please show the steps.
Hint: $(x + (1/x))\times (x^2 + (1/x^2)) = (x^3) + x + 1/x + 1/x^3$
We have : 
$(x + 1/x )^3$ 
= $x^3 + 3x^2(1 / x) + 3x(1 / x)^2 + (1/x)^3$
= $(x^3 + (1/x)^3 )  + 3x + 3/x$ 
= $(x^3 + (1/x)^3 ) + 3(x+1/x)$
=>
$(x^3 + (1/x)^3 ) = (x+1/x)^3 - 3(x+1/x) = 3^3 - 3.3 = 18.$