Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

a - b = 4, a³ - b³ = 88
then a² - b² = ?
(a - b)³ = a³ - b³ - 3ab(a - b)
64 = 88 - 12(ab)
ab = 2
(a + b)² - (a - b)² = 4ab
(a + b)² - 16 = 4 × 2
(a + b)² = 24
(a + b) = 2√6
a² - b² = (a + b) × (a - b) = 2√6 × 4 = 8√6

x³ + y³ + z³ = 3(1 + xyz)
P = y + z - x, Q = z + x - y, R = x + y - z
Put y and z = 0
x³ = 3, P = -x, Q = x, R = x
P³ + Q³ + R³ - 3PQR
= (-x)³ + (x)³ + (x)³ - 3(-x)(x)(x)
= x³ + 3x³
= 4x³
= 4 × 3
= 12

$$\eqalign{ & \because \,ab\left( {a + b} \right) = 1 \cr & \Rightarrow a + b = \frac{1}{{ab}}\,.......\,\left( {\text{i}} \right) \cr & {\text{On cubing both sides}} \cr & \Rightarrow {a^3} + {b^3} + 3ab\left( {a + b} \right) = \frac{1}{{{a^3}{b^3}}} \cr & \Rightarrow {a^3} + {b^3} + 3ab \times \frac{1}{{ab}} = \frac{1}{{{a^3}{b^3}}}\left( {{\text{from equation }}\left( {\text{i}} \right)} \right) \cr & \Rightarrow \frac{1}{{{a^3}{b^3}}} - {a^3} - {b^3} = 3 \cr} $$

ab(a - b) + bc(b - c) + ca(c - a)
Let c = 0
ab(a - b)
Now from option C
(b - a)(b - c)(c - a)
= -(a-b)b(-a)
= ab(a - b)

$$\eqalign{ & \frac{{\left( {a + b} \right)}}{c} = \frac{6}{5} \cr & \frac{{\left( {b + c} \right)}}{a} = \frac{9}{2} \cr & \frac{{\left( {a + c} \right)}}{b} = ? \cr & c = 5,\,\frac{{\left( {b + c} \right)}}{a} = \frac{9}{2} \cr & \Rightarrow \frac{{b + 5}}{2} = \frac{9}{2} \cr & b = 4,\,a = 2 \cr & \Rightarrow \frac{{\left( {a + c} \right)}}{b} = \frac{{2 + 5}}{4} \cr & \Rightarrow \frac{{\left( {a + c} \right)}}{b} = \frac{7}{4} \cr} $$

$$\eqalign{ & {\text{Given:}} \cr & x + \frac{{{y^2}}}{x} = 5 \cr & {\text{Put }}y = 2,\,x = 1 \cr & {\text{Now,}} \cr & \frac{{{x^2} + 2x + {y^2}}}{{{x^3} - 5{x^2}}} \cr & = \frac{{{{\left( 1 \right)}^2} + 2 \times 1 + {{\left( 2 \right)}^2}}}{{{{\left( 1 \right)}^3} - 5{{\left( 1 \right)}^2}}} \cr & = \frac{7}{{ - 4}} \cr & {\text{Put }}y = 2{\text{ in option and option D will satisfy the condition}} \cr & \cr & {\bf{Alternate \,solution:}} \cr & x + \frac{{{y^2}}}{x} = 5 \cr & {x^2} + {y^2} = 5x \cr & {x^2} - 5x = - {y^2} \cr & {\text{Now, }}\frac{{{x^2} + 2x + {y^2}}}{{{x^3} - 5{x^2}}} \cr & = \frac{{5x + 2x}}{{x\left( {{x^2} - 5x} \right)}} \cr & = \frac{{7x}}{{x\left( { - {y^2}} \right)}} \cr & = - \frac{7}{{{y^2}}} \cr} $$

a - b = 18, a³ - b³ = 324
(a - b)(a² + b² + ab) = 324
⇒ 18[(a - b)² + 3ab] = 324
⇒ 324 + 3ab = 18
⇒ 3ab = 18 - 324
⇒ 3ab = -306
⇒ ab = -102

(x - y) = 3
(x - y)³ = 27
⇒ x³ - y³ - 3xy(x - y) = 27
⇒ x³ - y³ - 9xy = 27

$$\eqalign{ & \Rightarrow \left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right) \cr & \Rightarrow \frac{{\left[ {\left( {x - \frac{1}{x}} \right)\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)} \right]}}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{{\left[ {\left( {{x^2} - \frac{1}{{{x^2}}}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)} \right]}}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{{\left[ {\left( {{x^4} - \frac{1}{{{x^4}}}} \right)\left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)} \right]}}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{{\left[ {\left( {{x^8} - \frac{1}{{{x^8}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)} \right]}}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{{\left[ {\left( {{x^{16}} - \frac{1}{{{x^{16}}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)} \right]}}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{{{x^{32}} - \frac{1}{{{x^{32}}}}}}{{x - \frac{1}{x}}} \cr} $$

$$\eqalign{ & {\text{A}} = \frac{{{{\left( {0.1} \right)}^3} + {{\left( {0.2} \right)}^3} + {{\left( {0.3} \right)}^3} + 3\left( {0.005 + 0.016 + 0.027} \right) + 0.036}}{{{{\left( {0.1} \right)}^2} + {{\left( {0.2} \right)}^2} + {{\left( {0.3} \right)}^2} + 0.04 + 0.06 + 0.12}} \cr & {\text{A}} = \frac{{{{10}^{ - 3}}\left[ {{1^3} + {2^3} + {3^3} + 3\left( {5 + 16 + 27} \right) + 36} \right]}}{{{{10}^{ - 2}}\left[ {{1^2} + {2^2} + {3^2} + 4 + 6 + 12} \right]}} \cr & {\text{A}} = \frac{{1 + 8 + 27 + 144 + 36}}{{10 \times 36}} \cr & {\text{A}} = \frac{{216}}{{10 \times 36}} \cr & {\text{A}} = \frac{6}{{10}} \cr & {\text{Hence, 60A}} = 60 \times \frac{6}{{10}} = 36{\text{ Answer}} \cr} $$