Question

if 3x-4=3/x then find the value of x^3-1/x^3

Answer 1

Answer:

$\frac{172}{27}$

Step-by-step explanation:

$3x-4=\frac{3}{x}\\3x - \frac{3}{x} = 4\\3(x-\frac{1}{x}) = 4\\x-\frac{1}{x}=\frac{4}{3}$

Squaring both sides

$(x - \frac{1}{x})^{2} = (\frac{4}{3})^{2}\\x^{2} +\frac{1}{x^{2} }-2= \frac{16}{9}\\x^{2} +\frac{1}{x^{2} }= \frac{16}{9}+2\\x^{2} +\frac{1}{x^{2} }= \frac{16+18}{9}\\x^{2} +\frac{1}{x^{2} }= \frac{34}{9}$

Now,

$a^{3} - b^{3} = (a-b)(a^{2}+b^{2}+ab)\\ x^{3}-\frac{1}{x^{3} } = (x-\frac{1}{x})[x^{2}+\frac{1}{x^{2} } +(x^{2})(\frac{1}{x^{2} })]\\ =(\frac{4}{3})(\frac{34}{9}+1)\\=(\frac{4}{3})(\frac{34+9}{9})\\=(\frac{4}{3})(\frac{43}{9})\\=\frac{172}{27}$