Question

If x = 2+√3 find the value of x³+1/x³​

Answer 1

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ value \ of \ x^{3}+\dfrac{1}{x^{3}} \ is \ 52.}$

$\sf\orange{Given:}$

$\sf{\leadsto{x=2+\sqrt3}}$

$\sf\pink{To \ find:}$

$\sf{The \ value \ of \ x^{3}+\dfrac{1}{x^{3}}.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{\longmapsto{x=2+\sqrt3}}$

$\sf{Taking \ cube \ of \ both \ sides, \ we \ get}$

$\sf{\longmapsto{\therefore{x^{3}=(2+\sqrt3)^{3}}}}$

$\sf{\longmapsto{\therefore{x^{3}=8+3\sqrt3+12\sqrt3+18}}}$

$\sf{\longmapsto{\therefore{x^{3}=26+15\sqrt3...(1)}}}$

$\sf{\longmapsto{\dfrac{1}{x}=\dfrac{1}{2+\sqrt3}}}$

$\sf{\longmapsto{\therefore{\dfrac{1}{x}=\dfrac{(1)(2-\sqrt3)}{(2+\sqrt3)(2-\sqrt3)}}}}$

$\sf{\longmapsto{\therefore{\dfrac{1}{x}=\dfrac{2-\sqrt3}{2^{2}-(\sqrt3)^{2}}}}}$

$\sf{\longmapsto{\therefore{\dfrac{1}{x}=\dfrac{2-\sqrt3}{4-3}}}}$

$\sf{\longmapsto{\therefore{\dfrac{1}{x}=\dfrac{2-\sqrt3}{1}}}}$

$\sf{\longmapsto{\therefore{\dfrac{1}{x}=2-\sqrt3}}}$

$\sf{On \ taking \ cube \ of \ both \ sides, \ we \ get}$

$\sf{\longmapsto{\therefore{\dfrac{1^{3}}{x^{3}}=(2-\sqrt3)^{3}}}}$

$\sf{\longmapsto{\therefore{\dfrac{1}{x^{3}}=8-12\sqrt3+18-3\sqrt3}}}$

$\sf{\longmapsto{\therefore{\dfrac{1}{x^{3}}=26-15\sqrt3...(2)}}}$

$\sf{...from \ (1) \ and \ (2)}$

$\sf{\longmapsto{x^{3}+\dfrac{1}{x^{3}}=(26+15\sqrt3)+(26-15\sqrt3)}}$

$\sf{\longmapsto{\therefore{x^{3}+\dfrac{1}{x^{3}}=52}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ x^{3}+\dfrac{1}{x^{3}} \ is \ 52.}}}$