A) \[\frac{P({{c}^{3}}+{{a}^{3}}+{{b}^{3}})}{4({{a}^{2}}+{{b}^{2}}-{{c}^{2}})}\]
B) \[\frac{P({{c}^{3}}-{{a}^{3}}-{{b}^{3}})}{4({{a}^{2}}+{{b}^{2}}-{{c}^{2}})}\]
C) \[P{{c}^{3}}-4{{a}^{2}}-4{{b}^{2}}\]
D) \[P{{c}^{3}}-2{{a}^{2}}-3{{b}^{2}}\]
Correct Answer: B
Solution :
[b] \[\left( P+\frac{4\sigma }{a} \right)\left( \frac{4}{3}\pi {{a}^{3}} \right)+\left( P+\frac{4\sigma }{b} \right)\left( \frac{4}{3}\pi {{b}^{3}} \right)\] \[=\left( P+\frac{4\sigma }{c} \right)\left( \frac{4}{3}\pi {{c}^{3}} \right)\] \[or\,\,\,P[{{a}^{3}}+{{b}^{3}}-{{c}^{3}}]=4\sigma [{{c}^{2}}-{{a}^{2}}-{{b}^{2}}]\] or \[\sigma =\frac{P({{c}^{3}}-{{a}^{3}}-{{b}^{3}})}{4({{a}^{2}}+{{b}^{2}}-{{c}^{2}})}\]
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