Question

pls answer this question​

Answer 1

$\large\underline{\underline{\mathfrak{\red{\bf{SOLUTION:-}}}}}$

$\bold{\underline{\:Find\:Here:-}}$

• Valiue of x.

$\large\underline{\underline{\mathfrak{\red{\bf{EXPLANATION-}}}}}$

$\bold{\boxed{\boxed{\:\frac{(x^2-x+1)}{(x^2+x+1)}\:=\frac{14(x-1)}{13(x+1)}}}}$

$\implies\:\frac{(x^2-x+1)(x+1)}{(x^2+x+1)(x-1)}\:=\frac{14}{13}$

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$\large\red{\underline{\:Using\:identity}}$

★$\red{\:(x+1)^3\:=\:(x+1)(x^2-x+1)}$

★$\red{\:(x-1)^3\:=\:(x-1)(x^2+x+1)}$

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$\implies\frac{(x+1)^3}{(x-1)^3}\:=\frac{14}{13}$

$\implies\left(\frac{(x+1)}{(x-1)}\right)^3\:=\frac{14}{13}$

$\implies\frac{(x+1)}{(x-1)}\:=\sqrt[3]{\frac{14}{13}}$

$\implies\sqrt[3]{{13}}(x+1)\:=\sqrt[3]{{14}}(x-1)$

$\implies\:x(\sqrt[3]{{14}}\:-\sqrt[3]{{13}})\:=\:(\sqrt[3]{{14}}\:-\sqrt[3]{{13}})$

$\implies\:x\:=\cancel{\frac{(\sqrt[3]{{14}}\:-\sqrt[3]{{13}})}{(\sqrt[3]{{14}}\:-\sqrt[3]{{13}}}}$

$\red{\bold{\boxed{\boxed{\:x\:=\:1}}}}$

$\large\underline{\underline{\mathfrak{\green{\bf{ANSWER:-}}}}}$

• Value of x = 1.