Question

If x+1/x=5 then find x^6-1/x^6

Answer 1

Answer: The value of $x^6-\frac{1}{x^6}$ is $2640\sqrt{21}$.

Explanation:

It is given that,

$x+\frac{1}{x}=5$

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

$(5)^2=x^2+\frac{1}{x^2}+2$

$x^2+\frac{1}{x^2}=23$

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

$(x-\frac{1}{x})^2=23-2$

$(x-\frac{1}{x})=\sqrt{21}$

Now use the formulas,

$a^3+b^3=(a+b)(a^2+b^2-ab)$

We get,

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

$x^3+\frac{1}{x^3}=(5)(23-1)$

$x^3+\frac{1}{x^3}=110$

Using $a^3-b^3=(a-b)(a^2+b^2+ab)$, we get

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

$x^3+\frac{1}{x^3}=(\sqrt{21})(23+1)$

$x^3+\frac{1}{x^3}=24\sqrt{21}$

Now using $a^2-b^2=(a-b)(a+b)$, we get,

$x^6+\frac{1}{x^6}=(x^3+\frac{1}{x^3})(x^3-\frac{1}{x^3})$

$x^6+\frac{1}{x^6}=110\times 24\sqrt{21}$

$x^6+\frac{1}{x^6}=2640\sqrt{21}$

Therefore, the value of $x^6-\frac{1}{x^6}$ is $2640\sqrt{21}$.