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
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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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).