Answer: Option B. -> 0
$$\eqalign{
& a = \frac{{{b^2}}}{{b - a}} \cr
& \Rightarrow a\left( {b - a} \right) = {b^2} \cr
& \Rightarrow ab - {a^2} = {b^2} \cr
& \Rightarrow {a^2} + {b^2} - ab = 0 \cr
& \Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right) \cr
& \Rightarrow {a^3} + {b^3} = 0 \cr} $$

Answer: Option C. -> -2
$$\frac{1}{{{x^{99}}}}{\text{ + }}\frac{1}{{{x^{98}}}} + \frac{1}{{{x^{97}}}} + \frac{1}{{{x^{96}}}} + \frac{1}{{{x^{95}}}} + \frac{1}{{{x^{94}}}} + \frac{1}{x} - 1$$
  $$ = \frac{1}{{{{\left( { - 1} \right)}^{99}}}}{\text{ + }}\frac{1}{{{{\left( { - 1} \right)}^{98}}}} + \frac{1}{{{{\left( { - 1} \right)}^{97}}}} + \frac{1}{{{{\left( { - 1} \right)}^{96}}}} + $$         $$\frac{1}{{{{\left( { - 1} \right)}^{95}}}} + $$   $$\frac{1}{{{{\left( { - 1} \right)}^{94}}}} + $$   $$\frac{1}{{\left( { - 1} \right)}} - $$   $$1$$
$$\eqalign{
& = - 1 + 1 - 1 + 1 - 1 + 1 + \frac{1}{{ - 1}} - 1 \cr
& = - 2 \cr} $$

Answer: Option A. -> 0
$$\eqalign{
& \frac{1}{{\root 3 \of 4 + \root 3 \of 2 + 1}} = a\root 3 \of 4 + b\root 3 \of 2 + c \cr
& \Rightarrow \frac{1}{{\root 3 \of 4 + \root 3 \of 2 + 1}} = \frac{1}{{{{\left( {{2^{\frac{1}{3}}}} \right)}^2} + {2^{\frac{1}{3}}} + {{\left( 1 \right)}^2}}} \cr
& \Rightarrow \therefore {{\text{A}}^3} - {{\text{B}}^3}{\text{ = }}\left( {{\text{A}} - {\text{B}}} \right)\left( {{{\text{A}}^2} + {\text{AB}} + {{\text{B}}^2}} \right) \cr
& {\text{Put, A}} = {2^{\frac{1}{3}}}{\text{, B}} = 1 \cr
& \Rightarrow \frac{{\left( {{2^{\frac{1}{3}}} - 1} \right)}}{{\left( {{2^{\frac{1}{3}}} - 1} \right)\left( {{{\left( {{2^{\frac{1}{3}}}} \right)}^2} + {2^{\frac{1}{3}}} + {{\left( 1 \right)}^2}} \right)}} \cr
& \Rightarrow \frac{{\left( {{2^{\frac{1}{3}}} - 1} \right)}}{{{{\left( {{2^{\frac{1}{3}}}} \right)}^3} - {{\left( 1 \right)}^3}}} \cr
& \Rightarrow \left( {{2^{\frac{1}{3}}} - 1} \right) \cr
& \therefore {2^{\frac{1}{3}}} - 1 \cr
& = a\left( {{2^{\frac{2}{3}}}} \right) + b{\left( 2 \right)^{\frac{1}{3}}} + c \cr
& \left( {{\text{Comparing the terms}}} \right) \cr
& a = 0 \cr
& b = 1 \cr
& c = - 1 \cr
& \therefore a + b + c \cr
& = 0 + 1 - 1 \cr
& = 0 \cr} $$

Answer: Option D. -> 4
$$\eqalign{
& {\text{ }}x = \root 3 \of {2 + \sqrt 3 } \cr
& {x^3} = 2 + \sqrt 3 \cr
& \frac{1}{{{x^3}}} = \frac{1}{{2 + \sqrt 3 }} \times \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }} = 2 - \sqrt 3 \cr
& \therefore {x^3}{\text{ + }}\frac{1}{{{x^3}}} \cr
& = 2 + \sqrt 3 + 2 - \sqrt 3 \cr
& = 4 \cr} $$

Answer: Option B. -> $$\frac{1}{{2{p^2}}}$$
$$\frac{{{p^2} - p}}{{2{p^3} + {p^2}}} + \frac{{{p^2} - 1}}{{{p^2} + 3p}} + \frac{{{p^2}}}{{p + 1}}$$
In such type of question assume values of p
$$\eqalign{
& \therefore {\text{Let }}p = 1 \cr
& \therefore \frac{{1 - 1}}{{2 + 1}} + \frac{{1 - 1}}{{1 + 3}} + \frac{1}{{1 + 1}} \cr
& = 0 + 0 + \frac{1}{2} \cr
& = \frac{1}{2} \cr
& {\text{Now check option 'B'}} \cr
& \frac{1}{{2{p^2}}} = \frac{1}{2} \cr
& {\text{Hence the answer is option 'B'}} \cr} $$

Answer: Option A. -> -7
$$\eqalign{
& {\text{ }}{x^2}{\text{ + }}\alpha x + \beta = 0 \cr
& {\text{Sum of root}} \cr
& \alpha + \beta = \frac{{ - \alpha }}{1}\,......(i) \cr
& \alpha \beta = \beta \,......(ii) \cr
& {\text{From (i) and (ii)}} \cr
& {\text{Then, }}\alpha = 1 \cr
& {\text{Then, }}\beta = - 2 \cr
& {\text{Then value of }} \cr
& \Leftrightarrow {\alpha ^3} + {\beta ^3} \cr
& = 1 + {\left( { - 2} \right)^3} \cr
& = - 7 \cr} $$

Answer: Option B. -> 5
$$\eqalign{
& {x^2} + \frac{1}{{{x^2}}} = 1 \cr
& {\text{Then, }}{\left( {x + \frac{1}{x}} \right)^2} = 1 + 2 \cr
& \Rightarrow x + \frac{1}{x} = \sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} \cr
& = {\left( {\sqrt 3 } \right)^3} - 3\sqrt 3 \cr
& = 3\sqrt 3 - 3\sqrt 3 \cr
& = 0 \cr
& {\text{Then,}} \cr
& {x^{102}} + {x^{96}} + {x^{90}} + {x^{84}} + {x^{78}} + {x^{72}} + 5 \cr} $$
  $$ = {x^{96}}\left( {{x^6} + 1} \right) + $$    $${x^{84}}\left( {{x^6} + 1} \right) + $$   $${x^{72}}\left( {{x^6} + 1} \right) + $$   $$5$$
  $$ = 5$$

Answer: Option C. -> 1
$$\eqalign{
& {x^3} - 7{x^2} + 11x - 5 \geqslant 0 \cr
& \Rightarrow {x^3} - 5{x^2} - 2{x^2} + 10x + x - 5 \geqslant 0 \cr
& \Rightarrow {x^2}\left( {x - 5} \right) - 2x\left( {x - 5} \right) + 1\left( {x - 5} \right) \geqslant 0 \cr
& \Rightarrow \left( {x - 5} \right)\left( {{x^2} - 2x + 1} \right) \geqslant 0 \cr
& \Rightarrow \left( {x - 5} \right){\left( {x - 1} \right)^2} \geqslant 0 \cr
& \Rightarrow \left( {x - 5} \right)\left( {x - 1} \right)\left( {x - 1} \right) \geqslant 0 \cr
& {\text{So, }}x = 1\& 5 \cr} $$
Equation satisfies at both the values, but the minimum value of these two
x = 1

Answer: Option C. -> -2, 2
If (x - 1) and (x + 1) are the factors y equation then,
$$\eqalign{
& x - 1 = 0 \cr
& x = 1 \cr
& \Rightarrow {\text{Put }}x = 1{\text{, we get }} \cr
& 1 + a - 3 + 2 + b = 0 \cr
& a + b = 0\,.....(i) \cr
& \Rightarrow x + 1 = 0 \cr
& \Rightarrow x = - 1 \cr
& {\text{Put }}x = - 1,{\text{we get }} \cr
& 1 - a - 3 - 2 + b = 0 \cr
& b - a = 4\,.....(ii) \cr
& {\text{After solving (i) & (ii),}} \cr
& {\text{We get }} \cr
& a = - 2,{\text{ }}b = 2 \cr} $$

Answer: Option A. -> 458
$$\eqalign{
& {\left( {a + b + c} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) \cr
& \Rightarrow {\left( {26} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {109} \right) \cr
& \Rightarrow {a^2} + {b^2} + {c^2} = 676 - 218 \cr
& \Rightarrow {a^2} + {b^2} + {c^2} = 458 \cr} $$