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Question

With usual notation in a ABC, if (1r1+1r2)(1r2+1r3)(1r3+1r1)=KR3a2b2c2, then K has the value equal to

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Solution

We know that,
r1=sa, r2=sb, r3=sc
Now, (1r1+1r2)(1r2+1r3)(1r3+1r1)=(sa+sb)(sb+sc)(sc+sa)=(2s(a+b))(2s(b+c))(2s(a+c))=cab=abc3 (1)
R=abc4=abc4R
Putting the value of in equation (1),
abc(abc)3×(4R)3=64R3a2b2c2=KR3a2b2c2
K=64

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