Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & {\text{ }}2x + \frac{2}{x} = 3 \cr & \Rightarrow {\text{ }}x + \frac{1}{x} = \frac{3}{2} \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( {\frac{3}{2}} \right)^3}{\text{ }} \cr & x + \frac{1}{x} = a \cr & {x^3} + \frac{1}{{{x^3}}} = {a^3} - 3a \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\left( {x + \frac{1}{x}} \right) = \frac{{27}}{8} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times \frac{3}{2} = \frac{{27}}{8} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = \frac{{27}}{8} - \frac{9}{2} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = \frac{{ - 9}}{8} \cr & \therefore {x^3} + \frac{1}{{{x^3}}} + 2 \cr & = \frac{{ - 9}}{8} + 2 \cr & = \frac{{ - 9 + 16}}{8} \cr & = \frac{7}{8} \cr} $$

$$\eqalign{ & a + b + c = 15{\text{ }} \cr & {a^2} + {b^2} + {c^2} = 83{\text{ }}\left( {{\text{Given}}} \right) \cr & \therefore a + b + c = 15 \cr & \left( {{\text{Squaring both sides}}} \right){\text{ }} \cr & \Rightarrow {\left( {a + b + c} \right)^2} = {\left( {15} \right)^2}{\text{ }} \cr & \Rightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = 225 \cr & \Rightarrow 83 + 2\left( {ab + bc + ca} \right) = 225 \cr & \Rightarrow 2\left( {ab + bc + ca} \right) = 142 \cr & \Rightarrow ab + bc + ca = 71 \cr} $$
  $$\therefore {a^3} + {b^3} + {c^3} - 3abc = $$     $$\left( {a + b + c} \right)$$  $$\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$
$$\eqalign{ & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 15\left( {83 - 71} \right) \cr & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 15 \times 12 \cr & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 180 \cr} $$

$$\eqalign{ & x + \frac{1}{{x + 1}} = 1 \cr & {\text{Adding both 1 sides}} \cr & \Rightarrow x + 1 + \frac{1}{{x + 1}} = 1 + 1 \cr & \Rightarrow \left( {x + 1} \right) + \frac{1}{{\left( {x + 1} \right)}} = 2 \cr & {\text{Put }}x + 1 = 1 \cr & {\text{And }}\frac{1}{{x + 1}} = 1 \cr & \therefore {\left( {x + 1} \right)^5}{\text{ + }}\frac{1}{{{{\left( {x + 1} \right)}^5}}} \cr & = 1 + 1 \cr & = 2 \cr} $$

$$\eqalign{ & {\text{If }}a + b + c = 0 \cr & {\text{then,}} \cr & {a^3} + {b^3} + {c^3} - 3abc = 0 \cr & \Leftrightarrow {a^3} + {b^3} + {c^3} = 3abc \cr} $$

x = y = 333,     z = 334
⇒ x³ + y³ + z³ - 3xyz = $$\frac{1}{2}$$ (x + y + z) [(x - y)² + (y - z)² + (z - x)²]
⇒ x³ + y³ + z³ - 3xyz = $$\frac{1}{2}$$ (333 + 333 + 334) (333 - 333)² + (333 - 334)² + (334 - 333)²
⇒ x³ + y³ + z³ - 3xyz = $$\frac{1}{2}$$ (1000) (0 + 1 + 1)
⇒ x³ + y³ + z³ - 3xyz = 1000

$$\eqalign{ & a = \frac{{{b^2}}}{{b - a}} \cr & \Rightarrow a\left( {b - a} \right) = {b^2} \cr & \Rightarrow ab - {a^2} = {b^2} \cr & \Rightarrow {a^2} + {b^2} - ab = 0 \cr & \Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right) \cr & \Rightarrow {a^3} + {b^3} = 0 \cr} $$

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

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

$$\eqalign{ & {\text{ }}x = \root 3 \of {2 + \sqrt 3 } \cr & {x^3} = 2 + \sqrt 3 \cr & \frac{1}{{{x^3}}} = \frac{1}{{2 + \sqrt 3 }} \times \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }} = 2 - \sqrt 3 \cr & \therefore {x^3}{\text{ + }}\frac{1}{{{x^3}}} \cr & = 2 + \sqrt 3 + 2 - \sqrt 3 \cr & = 4 \cr} $$

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