Question

if x^3/2 + 1/x^3/2 = 2 root5 then find the value of x^3 - 1/x^3 ..answer fast i'll mark u as brainliest

Answer 1

Answer:

Answer:

The value of \bold{x^{3}-\frac{1}{x}^{3}=215.47}x3−x13=215.47

Step-by-step explanation:

To find: Value of x^{3}-\frac{1}{x^{3}}x3−x31

According to formulae,

\begin{gathered}\begin{array}{l}\bold{{(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)}} \\ {a=x \text { and } b=\frac{1}{x}} \\ {\left(x-\frac{1}{x}\right)^{3}=x^{3}-\frac{1}{x}^{3}-3 . x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right)}\end{array}\end{gathered}(a−b)3=a3−b3−3ab(a−b)a=x and b=x1(x−x1)3=x3−x13−3.x⋅x1(x−x1)

As the question says  

x-\frac{1}{x}=3+2 \sqrt{2}x−x1=3+22

Now applying the formula, we get,

(3+2 \sqrt{2})^{3}=x^{3}-\frac{1}{x}^{3}-3(3+2 \sqrt{2})(3+22)3=x3−x13−3(3+22)

Solving the equation,

\begin{gathered}\begin{array}{l}{(3+2 \sqrt{2})^{3}+3(3+2 \sqrt{2})=x^{3}-\frac{1^{3}}{x}} \\ {195.112+17.48=x^{3}-\frac{1^{3}}{x}} \\ {215.47=\mathrm{x}^{3}-\frac{1^{3}}{\mathrm{x}}}\end{array}\end{gathered}(3+22)3+3(3+22)=x3−x13195.112+17.48=x3−x13215.47=x3−x13

Thus the value of x^{3}-\frac{1}{x}^{3}=\bold{215.47}x3−x13=215.47