| Statement I: \[\frac{{{({{a}^{2}}-{{b}^{2}})}^{3}}+{{({{b}^{2}}-{{c}^{2}})}^{3}}+{{({{c}^{2}}-{{a}^{2}})}^{3}}}{{{(a+b)}^{3}}{{(b+c)}^{3}}+{{(c+a)}^{3}}}\] |
| Statement II: \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca\] \[=\frac{1}{2}\left[ {{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}} \right]\] |
A) Both Statement-I and Statement-II are true.
B) Statement-I is true but Statement-II is false.
C) Statement-I is false but Statement-II is true.
D) Both Statement-I and Statement-II are false.
Correct Answer: C
Solution :
Statement - I: We have, \[\frac{{{({{a}^{2}}-{{b}^{2}})}^{3}}+({{b}^{2}}-{{c}^{2}})+{{({{c}^{2}}-{{a}^{2}})}^{3}}}{{{(a+b)}^{3}}+{{(b+c)}^{3}}+{{(c+a)}^{3}}}\] \[\frac{\begin{align} & ({{a}^{2}}-{{b}^{2}}+{{b}^{2}}-{{c}^{2}}+{{c}^{2}}-{{a}^{2}})[{{({{a}^{2}}-{{b}^{2}})}^{2}}+{{({{b}^{2}}-{{c}^{2}})}^{2}} \\ & +{{({{c}^{2}}-{{a}^{2}})}^{2}}-({{a}^{2}}-{{b}^{2}})({{b}^{2}}-{{c}^{2}})-({{b}^{2}}-{{c}^{2}})({{c}^{2}}-{{a}^{2}}) \\ & -{{({{c}^{2}}-{{a}^{2}})}^{2}}({{a}^{2}}-{{b}^{2}})]+3({{a}^{2}}-{{b}^{2}})({{b}^{2}}-{{c}^{2}})({{c}^{2}}-{{a}^{2}}) \\ \end{align}}{\begin{align} & (a+b+b+c+c+a)[{{(a+b)}^{2}}+{{(b+c)}^{2}}+{{(c+a)}^{2}} \\ & -(a+b)(b+c)-(b+c)(c+a)-(c+a) \\ & (a+b)]+3(a+b)(b+c)(c+a) \\ \end{align}}\] Statement -II: We have, \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac=\frac{1}{2}\{2a-2{{b}^{2}}\] \[+2{{c}^{2}}-2ab-2bc-2ac\}\] \[=\frac{1}{2}[{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}]\] \[\therefore \]Statement - I is false but Statement -II is true.
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