Question

x^2×1/25x^2=43/5 find the value of x^3+1/125x^3​

Answer 1

Answer: $\pm\dfrac{126}{5}$  depending if x is positive or negative

Step-by-step explanation:

First you should notice that

$\left(x+\dfrac{1}{5x}\right)^2=x^2+2x\cdot\dfrac{1}{5x}+\left(\dfrac{1}{5x}\right)^2=x^2+\dfrac{2}{5}+\dfrac{1}{25x^2}$

So,

$x^2+\dfrac{1}{25x^2}=\dfrac{43}{5}\Longrightarrow \left(x+\dfrac{1}{5x}\right)^2=\dfrac{43}{5}+\dfrac{2}{5}=\dfrac{45}{5}=9$

from here we have that

$x+\dfrac{1}{5x}=\pm3$

on the other hand, using the fact that $a^3+b^3=(a+b)(a^2-ab+b^2)$ then

$x^3+\dfrac{1}{125x^3}=x^3+\left(\dfrac{1}{5x}\right)^3=\left(x+\dfrac{1}{5x}\right)\left(x^2-x\cdot\dfrac{1}{5x}+\dfrac{1}{25x^2}\right)=\pm3\cdot\left(\dfrac{43}{5}-\dfrac{1}{5}\right)=\pm\dfrac{126}{5}$