Question

${a^2+ab+b^2=25}$
${b^2+bc+c^2=49}$
${c^2+ca+a^2=64}$
Then, find the value of
(a+b+c)²-100 = ????​

Answer 1

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).

❁❁ Refer To Image Now .. ❁❁

Answer 2

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

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 common factor  = ±3  or ±1

all three terms have a + b + c  as common

=> a + b + c = ±3 or ±1

a + b + c  = ±1 does not give integral values of a , b & c

a + b + c = ±3

a = 0  , b = 5 , c  = -8     or a = 0 , b = - 5 c = 8

=> (a + b + c) ² - 100

= (±3)² - 100

= 9 - 100

= - 91

(a + b + c) ² - 100 = - 91

a = 0  , b = 5 , c  = -8     or a = 0 , b = - 5 c = 8

Verification :

0  + 0 + 25  = 25  ,

25  - 40  + 64  = 49    ,

64  + 0 +  0  = 64

There is another solution  (a + b + c) ² - 100 = 29  if a , b , c ≠ 0 and a , b , c ∈ R

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