Question

if x^4-79x^2+1=0 then find the value of x^3+1/x^3

Answer 1

$\textbf{Given equation is}$

$x^4-79x^2+1=0$

$\text{This can be written as}$

$x^4+1=79x^2$

$\implies\bf\displaystyle\,x^2+\frac{1}{x^2}=79$

$\text{We know that,}$

$\boxed{\bf\,(a+b)^2=a^2+b^2+2ab}$

$\implies\displaystyle(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2x(\frac{1}{x})$

$\implies\displaystyle(x+\frac{1}{x})^2=79+2$

$\implies\displaystyle(x+\frac{1}{x})^2=81$

$\implies\bf\displaystyle\,x+\frac{1}{x}=9$

$\text{We know that,}$

$\boxed{\bf\,a^3+b^3=(a+b)^3-3ab(a+b)}$

$\implies\displaystyle\,x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3(x)(\frac{1}{x})(x+\frac{1}{x})$

$\implies\displaystyle\,x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3(x+\frac{1}{x})$

$\implies\displaystyle\,x^3+\frac{1}{x^3}=(9)^3-3(9)$

$\implies\displaystyle\,x^3+\frac{1}{x^3}=729-27$

$\implies\boxed{\bf\,x^3+\frac{1}{x^3}=702}$

$\therefore\textbf{The value of \bf\,x^3+\frac{1}{x^3} is 702}$