# DIFFERENT PROBLEMS USING ALGEBRAIC IDENTITIES

## About "Different problems using algebraic identities"

Different problems using algebraic identities :

Here we are going to see some different problems using identities.

Problem 1 :

If (x + a) (x + b) (x + c) = x³ - 10 x² + 45 x - 15 find a+b+c and 1/a+1/b+1/c and a² +  b² + c²

Solution :

(x + a)(x + b)(x + c) = x³ + (a+b+c)x² + (ab + bc + ca)x + abc

x³+(a+b+c)x²+(ab+bc+ca)x+abc = x³ - 10 x² + 45 x - 15

By comparing these two equations coefficients of x², x and constant are equal.

Coefficient of x² :

a + b + c  =  -10

Coefficient of x :

ab + bc + ca  =  45

Constant term :

abc =  -15

(i) a + b +c  =  -10

(ii)(1/a) + (1/b) + (1/c)

To find the value of (1/a) + (1/b) + (1/c), we have to find the least common multiple of denominators. =  45/(-15)

=  -3

(iii) a² +  b² + c²

To find the value of a² +  b² + c², we have to use the algebraic identity of (a + b + c)².

(a + b + c)²  =  a² + b² + c² + 2ab + 2bc + 2ca

(a + b + c)² =  a² + b² + c² + 2(ab + bc + ca)

a² + b² + c²  = (a + b + c)² - 2(ab + bc + ca)

=  (-10)² - 2 (45)

=  100 - 90

= 10

Problem 2 :

Find 8x³ + 27y³ if 2x + 3y = 13 and xy = 6

Solution :

8x³ + 27y³ =  2³x³ + 3³y³ ==>  (2x)³ + (3y)³

a³ + b³ = (a+b)³ - 3ab (a+b)

a = 2x and b = 3y

(2x)³ + (3y)³ = (2x + 3y)³ - 3(2x)(3y)(2x + 3y)

=  13³- 18xy (13)

=  2197- 18(6) (13)

=  2197- 1404

=  793

After having gone through the stuff given above, we hope that the students would have understood "Different problems using algebraic identities".

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