Question

if x is equal to 9 minus 4 root 5 find x cube minus one by x cube​

Answer 1

Answer:

CORRECT QUESTION

x = 9 - 4√5

Answers

THE BRAINLIEST 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}=5788