Question

solve for X : 1+x+x^2/1-x+x^2=171(1+x)/172(1-x)​

Answer 1

$\textbf{Given:}$

$\dfrac{1+x+x^2}{1-x+x^2}=\dfrac{171(1+x)}{172(1-x)}$

$\textbf{To find:}$

$\text{The value of x}$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{1+x+x^2}{1-x+x^2}=\dfrac{171(1+x)}{172(1-x)}$

$172(1-x)(1+x+x^2)=171(1+x)(1-x+x^2)$

$\text{Using, the following identities}$

$\boxed{\bf\,a^3+b^3=(a+b)(a^2-ab+b^2)}$

$\boxed{\bf\,a^3-b^3=(a-b)(a^2+ab+b^2)}$

$172(1^3-x^3)=171(1^3+x^3)$

$172(1-x^3)=171(1+x^3)$

$172-172\,x^3=171+171\,x^3$

$172-171=171\,x^3+172\,x^3$

$1=343\,x^3$

$x^3=\dfrac{1}{343}$

$x^3=(\frac{1}{7})^3$

$\implies\,x=\dfrac{1}{7}$

$\therefore\textbf{The value of x is \bf\dfrac{1}{7} }$

Find more:

Solve the following quadratic equations by factorization:(x+3/x-2)-(1-x/x)=17/4

https://brainly.in/question/15926221

(2x-3)=root2x^2-2x+21

https://brainly.in/question/6115327

Answer 2

Step-by-step explanation:

thanks for giving this type of questions