# USING THE sing the FORMULA OF EXPANSION OF BINOMIAL OF POWER 3

## About "Using the Formula of Expansion of Binomial of Power 3"

Using the Formula of Expansion of Binomial of Power 3

Here we are going to see some example problems to show how to expand cubes of binomial.

(a + b)3  =  a3 + 3a2b + 3ab2 + b3

(a - b)3  =  a3 - 3a2b + 3ab2 - b3

(a3 + b3)  =  (a + b)3 - 3ab(a + b)

(a3 - b3)  =  (a - b)3 - 3ab(a - b)

## Using the Formula of Expansion of Binomial of Power 3 - Examples

Question 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

=  -85

Question 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

Question 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))=  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.

Question 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.

After having gone through the stuff given above, we hope that the students would have understood, "Using the Formula of Expansion of Binomial of Power 3"

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