A) \[{{a}^{2}}+{{b}^{2}}\]
B) \[{{a}^{3}}+{{b}^{3}}\]
C) \[{{a}^{3}}{{b}^{3}}\]
D) \[{{a}^{3}}-{{b}^{3}}\]
Correct Answer: B
Solution :
If \[x=a+b,y=a\alpha +b\beta \] and \[z=\alpha \beta +b\alpha \] Then \[xyz=(a+b)(a\omega +b{{\omega }^{2}})(a{{\omega }^{2}}+b\omega ),\]where \[\alpha =\omega \] and \[\beta ={{\omega }^{2}}\] \[=(a+b)({{a}^{2}}+ab{{\omega }^{2}}+ab\omega +{{b}^{2}})\] \[=(a+b)({{a}^{2}}-ab+{{b}^{2}})={{a}^{3}}+{{b}^{3}}\] Trick: Put \[a=b=2\] then\[x=4,y=2(\omega +{{\omega }^{2}})=-2\]and \[z=2({{\omega }^{2}}+\omega )=-2\] \[\therefore \] \[xyz=4(-2)(-2)=16\]and (b) i.e. \[{{a}^{3}}+{{b}^{3}}=16\].
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