Question

if x=9-4√5 then find x^3+1/x^3

Answer 1

this is the solution

Answer 2

Answer:

$x^{3}+\frac{1}{x^3}=5778$

Step-by-step explanation:

We need to find $x^{3}+\frac{1}{x^3}$

If $x=9-4\sqrt{5}$

so, $\frac{1}{x}=\frac{1}{9-4\sqrt{5}}$

Rationalize the above

$\frac{1}{x}=\frac{1}{9-4\sqrt{5}}\times \frac{9+4\sqrt{5}}{9+4\sqrt{5}}$

$\frac{1}{x}=\frac{9+4\sqrt{5}}{9-4\sqrt{5}(9+4\sqrt{5})}$

$\frac{1}{x}=\frac{9+4\sqrt{5}}{81-80}$

$\frac{1}{x}=9+4\sqrt{5}$

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

Put $x=9-4\sqrt{5}$ in above

$(9-4\sqrt{5}+\frac{1}{9-4\sqrt{5}})^2=x^{2}+\frac{1}{x^2}+2$

$(9-4\sqrt{5}+9+4\sqrt{5})^2=x^{2}+\frac{1}{(9-4\sqrt{5})^2}+2$

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

$(18)^2=x^{2}+\frac{1}{x^2}+2$

$324=x^{2}+\frac{1}{x^2}+2$

subtract both the sides by 2,

$324-2=x^{2}+\frac{1}{x^2}$

$322=x^{2}+\frac{1}{x^2}$

Then

$x^{3}+\frac{1}{x^3}=(x+\frac{1}{x})(x^{2}+\frac{1}{x^2}-x\times \frac{1}{x})$

$x^{3}+\frac{1}{x^3}=(x+\frac{1}{x})(x^{2}+\frac{1}{x^2}-1)$

$x^{3}+\frac{1}{x^3}=(18)(322-1)$

$x^{3}+\frac{1}{x^3}=(18)(321)$

$x^{3}+\frac{1}{x^3}=5778$