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Algebra: Polynomials Test4
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Question 1 
$ \displaystyle If\,\,x+y=7,\,\,then\,\,the\,\,value\,\,\,of\,\,{{x}^{3}}+{{y}^{3}}+21xy\,\,\,is$
243
 
143  
343  
443 
Question 1 Explanation:
$ \displaystyle \begin{array}{l}Given,\,\,x+y=7\\Now,\,\,{{x}^{3}}+{{y}^{3}}+21xy\\={{\left( x+y \right)}^{3}}3xy\left( x+y \right)+21xy\\={{\left( 7 \right)}^{3}}3xy\,\left( 7 \right)+21xy\\34321xy+21xy=343\end{array}$
Question 2 
$ \displaystyle If\,\,\,{{x}^{\frac{1}{3}}}+{{y}^{\frac{1}{3}}}={{z}^{\frac{1}{3}}},\,\,then\,\,\,\left[ {{\left( x+yz \right)}^{3}}+27xyz \right]$
1  
1  
0  
27

Question 2 Explanation:
$ \displaystyle \begin{array}{l}{{x}^{\frac{1}{3}}}+{{y}^{\frac{1}{3}}}={{z}^{\frac{1}{3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,................\left( i \right)\\Cubing\,\,\,both\,\,\,sides,\,\\{{\left( {{x}^{\frac{1}{3}}}+{{y}^{\frac{1}{3}}} \right)}^{3}}=z\\\Rightarrow x+y+3\,\,\,{{x}^{\frac{1}{3}}}.y{{\,}^{\frac{1}{3}}}\,\,\left( {{x}^{\frac{1}{3}}}+{{y}^{\frac{1}{3}}} \right)=z\\=>x+yz=3.{{x}^{\frac{1}{3}}}.{{y}^{\frac{1}{3}}}.{{z}^{\frac{1}{3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...............\left( ii \right)\\=>putting\text{ }in\text{ }equation\\{{\left( x+yz \right)}^{3}}\,\,\,+27xyz=+27\,\,xyz27xyz=0\end{array}$
Question 3 
$ \displaystyle If\,\,\,x\frac{1}{x}=4,\,\,\,then\,\,\left( x+\frac{1}{x} \right)$
$ \displaystyle 5\sqrt{2}$  
$ \displaystyle 2\sqrt{5}$  
$ \displaystyle 4\sqrt{2}$  
$ \displaystyle 4\sqrt{5}$ 
Question 3 Explanation:
$ \begin{array}{l}{{\left( x\frac{1}{x} \right)}^{2}}=16~~~\\=>{{x}^{2}}+\frac{1}{{{x}^{2}}}=18+2=20\\=>x+\frac{1}{x}=2\sqrt{5}\end{array}$
Question 4 
$ \displaystyle If\,\,\,x=3+\sqrt{8,\,}\,\,then\,\,the\,\,value\,\,of\,\,\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)$
34  
24  
38  
36

Question 4 Explanation:
$ \begin{array}{l}x+\frac{1}{x}=3+\sqrt{8}+\frac{1}{3+\sqrt{8}}=3+\sqrt{8}+\frac{3\sqrt{8}}{\left( 3+\sqrt{8} \right)\left( 3\sqrt{8} \right)}=6\\=>{{x}^{2}}+\frac{1}{{{x}^{2}}}={{\left( x+\frac{1}{x} \right)}^{2}}2\\=>362=34\end{array}$
Question 5 
$ \displaystyle If\,\,\,4{{b}^{2}}+\frac{1}{{{b}^{2}}}=2,\,\,then\,\,\,the\,\,\,value\,\,of\,\,8{{b}^{3}}+\frac{1}{{{b}^{3}}}\,\,is$
0  
1  
2  
5 
Question 5 Explanation:
$ \displaystyle \begin{array}{l}{{\left( 2b+\frac{1}{b} \right)}^{2}}=4{{b}^{2}}+\frac{1}{{{b}^{2}}}+4=6\\\Rightarrow 2b+\frac{1}{b}=\sqrt{6}\\Therefore\,\,\,\,\,\\=>{{\left( 2b+\frac{1}{b} \right)}^{3}}=8{{b}^{3}}+\frac{1}{{{b}^{3}}}+3\times 2b\times \frac{1}{b}\left( 2b+\frac{1}{b} \right)\\=>\left( 6\sqrt{6} \right)=8{{b}^{3}}+\frac{1}{{{b}^{3}}}+6\sqrt{6}\\=>8{{b}^{3}}+\frac{1}{{{b}^{3}}}=0\end{array}$
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