Question

plz someone help me to do this no.​

Answer 1

☆Given

X-$\frac{1}{X}$=√2

To Find

(a)X²+$\frac{1}{X^2}$

X-$\frac{1}{X}$=√2(both side squaring)

=>x²+$\frac{1}{x^2}$-2=2

=>x²+$\frac{1}{x^2}$=2+2

=>x²+$\frac{1}{x^2}$=4

$\because$x²+$\frac{1}{x^2}$=4 ---(1)

Now finding for

X+$\frac{1}{X}$ (squaring it)

=x²+$\frac{1}{x^2}$+2

=x²+$\frac{1}{x^2}$+2(putting value of eq1)

=4+2

=6 -----(2)

Now finding the value for

x³+$\frac{1}{x^3}$

=>(X+$\frac{1}{X}$)³ (cubing it)

=> (X+$\frac{1}{X}$)³=x³+$\frac{1}{x^3}$+3×x×$\frac{1}{X}$ (X+$\frac{1}{X}$ ) (putting value of equ 2)

=>(X+$\frac{1}{X}$)³=x³+$\frac{1}{x^3}$+3(6)

=>(6)³=x³+$\frac{1}{x^3}$+3(6)

=>216=x³+$\frac{1}{x^3}$+18

=>216-18=x³+$\frac{1}{x^3}$

192=x³+$\frac{1}{x^3}$