Question

a2+ab+b2=25
{b^2+bc+c^2=49}b2+bc+c2=49
{c^2+ca+a^2=64}c2+ca+a2=64
Then, find the value of
(a+b+c)²-100 = ????
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Answer 1

Answer:

Here is a purely algebraic way. Subtract the equations in pairs to get

(C−B)(A+B+C)=28,

(C−A)(A+B+C)=13,

(A−B)(A+B+C)=15.

Let us write x=(A+B+C)2 and y=BC+CA+AB. Then squaring and adding the above three equations yields

x(x−3y)=589.

Summing the original equations gives

2x−3y=149.

Eliminating 3y between these equations results in the quadratic equation

x2−149x+589=0.

Solving this, we get

x=12(149±63√5),

where the larger root corresponds to the case when A, B, and C are all positive.

Answer 2

Answer:

Formula,

(a+b+c)

2

=a

2

+b

2

+c

2

+2ab+2bc+2ca

⇒(a+b+c)

2

=a

2

+b

2

+c

2

+2(ab+bc+ca)

Given,

ab+bc+ca=10,a

2

+b

2

+c

2

=16

⇒(a+b+c)

2

=16+2(10)

a+b+c=

16+20

=6