ALGEBRAIC FORMULA

লেখক :

November 26, 2014

426

a²+ $\frac{1}{a^{2}}$ = 14 then a³+ $\frac{1}{a^{3}}$ = ?

Hints : a³+ b³= ( a+b) ( a²– ab + b²)

a³+ b³= ( a+b) ( a²– ab + b²)

a³+ $\frac{1}{a^{3}}$ = ( a+ $\frac{1}{a}$ ) ( a²– a. $\frac{1}{a}$ + $\frac{1}{a^{2}}$ )

= ( a+ $\frac{1}{a}$ ) ( a²+ $\frac{1}{a^{2}}$ – 1 )

= ( a+ $\frac{1}{a}$ ) ( 14 – 1)

= 13 ( a+ $\frac{1}{a}$ ) …………………………………………………(i)

a²+ $\frac{1}{a^{2}}$ = 14

or, ( a + $\frac{1}{a}}$ )²– 2.a. $\frac{1}{a}}$ = 14

or, ( a + $\frac{1}{a}}$ )²– 2 = 14

or, ( a + $\frac{1}{a}}$ )²= 14+2

or, ( a + $\frac{1}{a}}$ )²= 16

or, ( a + $\frac{1}{a}}$ )²= 4²

or, ( a + $\frac{1}{a}}$ ) = 4

Put the value in (i) equation then we get,

a³+ $\frac{1}{a^{3}}$ =13 ( a+ $\frac{1}{a}$ )

= 13 . 4

= 52