USING THE FORMULA OF EXPANSION OF BINOMIAL OF POWER 3

Example 1 :

Find x³ - y³, if  x - y  =  5 and xy  =  14

Solution :

x³ - y³, if  x - y  =  5 and xy  =  14

(a³ - b³)  =  (a - b)³ + 3ab(a - b)

(x³ - y³)  =  (x - y)³ + 3xy(x - y)

By using the given values, we get

(x³ - y³)  =  5³ + 3(14)(5)

=  125 + 210

=  335

Example 2 :

If a + (1/a)  =  6, then find the value of a³ + 1/a³

Solution :

(a³ + b³)  =  (a + b)³ - 3ab(a + b)

a³ + (1/a)³  =   (a + (1/a))³ - 3a(1/a)(a + (1/a))

  =  6³ - 3(6)

  =  216 - 18

  =  198

Example 3 :

If x² + 1/x²  =  23, then find the value of x + (1/x) and x³ + (1/x³)

Solution :

a² + b²  =  (a + b)² - 2ab

x² + (1/x)²  =  (x + (1/x))² - 2x(1/x)

23  =  (x + (1/x))² - 2

23 + 2  =  (x + (1/x))²

 (x + (1/x))²  =  25

x + (1/x)  =  5

x³ + (1/x)³  =  (x + (1/x))³ - 3x(1/x)(x + (1/x))

  =  5³ - 3(5)

  =  125 - 15

  =  105

Hence the values of x + (1/x) and x³ + (1/x)³ are 5 and 105 respectively.

Example 4 :

If (y - (1/y))³  =  27, then find the value of y³ - (1/y)³

Solution :

(a³ - b³)  =  (a - b)³ - 3ab(a - b)

Given that :

(y - (1/y))³  =  27

(y - (1/y))³  =  3³

(y - (1/y))  =  3

(y³ - (1/y)³)  =  (y - (1/y))³ - 3y(1/y)(y - (1/y))

  =  27 - 3(3)

  =  27 - 9

  =  18

Hence the value of (y³ - (1/y)³)  is 18.

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