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