Question
If (a + b)2Â = 18 + c2, (b +
c)2Â = 22 + a2Â and (c + a)2Â = 24 + b2, then find the value of (a + b + c).Solution
(a + b)2 = 18 + c2 So, (a + b)2 – c2 = 18 Or, (a + b + c)(a + b – c) = 18 – [equation I] (b + c)2 = 22 + a2 (b + c)2 – a2 = 22 Or, (b + c + a)(b + c – a) = 22 – [equation II] (c + a)2 = 24 + b2 (c + a)2 – b2 = 24 Or, (c + a + b)(c + a – b) = 24 – [equation III] Adding equations I, II and III (a + b + c){(a + b – c) + (b + c – a) + (c + a – b)} = 18 + 22 + 24 (a + b + c) {(a + b + c)} = 64 Or, (a + b + c)2 = 64 So, (a + b + c) = ±8
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