Math, asked by ayesha7351, 11 months ago

please solve it fast guys....if a,b,c are real numbers.....a^2+ab+b^2=25,b^2+bc+c^2=49, c^2+ca+a^2=64, then find the value of (a+b+c)^2-100​

Answers

Answered by amitnrw
10

Given  :  a²  + ab + b²  = 25  , b²  + bc + c²  = 49    , c²  + ac + a²  = 64    Eq3

To find : (a + b + c) ² - 100

Solution:

Assumption a  , b , c are integers

a²  + ab + b²  = 25   Eq1

b²  + bc + c²  = 49    Eq2

c²  + ac + a²  = 64    Eq3

Eq2 - Eq1

=> c² - a²  +  bc - ab  = 49 - 25

=> (c + a)(c - a) + b(c - a) = 24

=> (c - a)(a + b + c) = 24

Eq3 - Eq1

=>  (c - b)(a + b + c)  = 39

Eq3 - Eq2

(a - b)(a + b + c)  = 15

(c - a)(a + b + c) = 24

(c - b)(a + b + c)  = 39

(a - b)(a + b + c)  = 15

24 ,  39 & 45

have HCF = ±3

all three terms have a + b + c  as common

=> a + b + c = ±3

(a + b + c) ² - 100

= (±3)² - 100

= 9 - 100

= - 91

(a + b + c) ² - 100 = - 91

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Answered by RvChaudharY50
41

Given :-

  • a^2+ab+b^2 = 25
  • b^2+bc+c^2 = 49
  • c^2+ca+a^2 = 64

To Find :-

  • (a + b + c)² - 100 = ?

Formula used :-

  • (x-y)(x² + y² + xy) = (x³ - y³)
  • (x² - y²) = (x + y)(x - y)
  • x⁴ + y⁴ + x²y² = (x² + y² + xy)(x² + y² - xy).

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