Question

विस्तार करा: $\bigg \lgroup x+\frac{1}{x} \bigg \rgroup^3$

Answer 1

simply use the formula of (a+b)³ to expand given equation ...

(a+b)³ = a³+b³+3ab(a+b)

= x³+1/x³ + 3×x×1/x(x+1/x/

= x³+1/x³+3(x²+1/x)

Solved !

Answer 2

विस्तार करा: $\bigg \lgroup x+\frac{1}{x} \bigg \rgroup^3$

विस्तार सूत्र:

${(a+b)}^{3} = {a}^{3} + {b}^{3} + 3 {a}^{2} b + 3a {b}^{2} \\ \\ a=x \\ \\ b=\frac{1}{x} \\ \\ {\bigg(x+\frac{1}{x}\bigg)}^{3} = {(x)}^{3} + {\bigg(\frac{1}{x}}\bigg)^{3} + 3 {(x)}^{2} \bigg(\frac{1}{x}\bigg)+ 3(x) {\bigg(\frac{1}{x}\bigg)}^{2} \\ \\ = {x}^{3}+ \frac{1}{x^{3}} + 3x+ \frac{3}{x}$

विस्तार:
$\\ {\bigg(x+\frac{1}{x}\bigg)}^{3} ={x}^{3}+ \frac{1}{x^{3}} + 3x+ \frac{3}{x}$