Question

If $(5x+\frac{1}{5x})=7$; find the value of $(125x^3+\frac{1}{125x^3})$.

Answer 1

Given Expression is $(5x+\frac{1}{5x})=7$ ....(1)

Now, the find cube of (1) by using the formula $(a+b{)}^3=a^3+b^3+3ab(a+b)$

$\Rightarrow {(5x+\frac{1}{5x})}^3=(5x{)}^3+{\left(\frac{1}{5x}\right)}^3+3(5x)\left(\frac{1}{5x}\right)(5x+\frac{1}{5x})$

$\Rightarrow {(5x+\frac{1}{5x})}^3=125x^3+\left(\frac{1}{125x^3}\right)+3(5x+\frac{1}{5x})$

$\therefore (125x^3+\frac{1}{125x^3})={(5x+\frac{1}{5x})}^3-3(5x+\frac{1}{5x})$

$=(7{)}^3-3\times 7$
$=343-21$
$=322$
Hence, the value of $(125x^3+\frac{1}{125x^3})$ is 322.