Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

x³ + y³ + z³ = (x + y + z)(x² + y² + z² - xy - yz - zx)
75 - 3 × 15 = 10(x² + y² + z² - xy - yz - zx)
x² + y² + z² - xy - yz - zx = 3

x³ + y³ + z³ - 3xyz = (x + y + z)[x² + y² + z² - (xy + yz + zx)]
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
289 = x² + y² + z² + 2 × 111
x² + y² + z² = 67
Now,
x³ + y³ + z³ - 3xyz = 17(67 - 111)
x³ + y³ + z³ - 3xyz = -44 × 17
x³ + y³ + z³ - 3xyz + 4xyz = -748 + 4xyz
x³ + y³ + z³ + xyz = -748 + 4 × 171
x³ + y³ + z³ + xyz = -748 + 684
x³ + y³ + z³ + xyz = -64
$$\eqalign{ & \root 3 \of {\left( {{x^3} + {y^3} + {z^3} + xyz} \right)} \cr & = \root 3 \of { - 64} \cr & = - 4 \cr} $$

x³ + y³ = (x + y)[(x + y)² - 3xy]
1064 = 14[14² - 3xy]
76 = 196 - 3xy
3xy = 120
xy = 40
(x - y)²
= (x + y)² - 4xy
= 14² - 4 × 40
= 196 - 160
= 36

$$\eqalign{ & p + \frac{1}{p} = 112,\,\,{\left( {p - 112} \right)^{15}} + \frac{1}{{{p^{15}}}} = ? \cr & p - 112 = - \frac{1}{p} \cr & {\left( {p - 112} \right)^{15}} + \frac{1}{{{p^{15}}}} \cr & = {\left( { - \frac{1}{p}} \right)^{15}} + \frac{1}{{{p^{15}}}} \cr & = - \frac{1}{{{p^{15}}}} + \frac{1}{{{p^{15}}}} \cr & = 0 \cr} $$

x = 1 + √2 + √3
⇒ x - 1 = √2 + √3
On squaring both sides,
(x - 1)² = (√2 + √3)²
⇒ x² - 2x + 1 = 2 + 3 + 2√6
Add 3 both side
⇒ x² - 2x + 4 = 5 + 2√6 + 3
⇒ x² - 2x + 4 = 8 + 2√6
⇒ x² - 2x + 4 = 2(4 + √6)

$$\eqalign{ & {a^2} + {b^2} + {c^2} + 96 = 8\left( {a + b - 2c} \right) \cr & \Rightarrow {a^2} + {b^2} + {c^2} + 96 = 8a + 8b - 16c \cr & \Rightarrow {a^2} + {b^2} + {c^2} + 96 - 8a - 8b + 16c = 0 \cr & \Rightarrow {a^2} - 8a + 16 + {b^2} - 8b + 16 + {c^2} + 16c + 64 = 0 \cr & \Rightarrow {a^2} - 2 \times a \times 4 + {4^2} + {b^2} - 2 \times b \times 4 + {4^2} + {c^2} + 2 \times c \times 8 + {8^2} = 0 \cr & \Rightarrow {\left( {a - 4} \right)^2} + {\left( {b - 4} \right)^2} + {\left( {c + 8} \right)^2} = 0 \cr & {\text{Now we can say,}} \cr & \Rightarrow {\left( {a - 4} \right)^2} = 0 \cr & \Rightarrow a - 4 = 0 \cr & \Rightarrow a = 4 \cr & {\text{Similarly,}} \cr & \Rightarrow {\left( {b - 4} \right)^2} = 0 \cr & \Rightarrow b - 4 = 0 \cr & \Rightarrow b = 4 \cr & {\text{Similarly,}} \cr & \Rightarrow {\left( {c + 8} \right)^2} = 0 \cr & \Rightarrow c + 8 = 0 \cr & \Rightarrow c = - 8 \cr & {\text{Now, put }}a,{\text{ }}b{\text{ and }}c{\text{ value in }}\sqrt {ab - bc - ca} \cr & \Rightarrow \sqrt {4 \times 4 - 4 \times \left( { - 8} \right) + \left( { - 8} \right) \times 4} \cr & \Rightarrow \sqrt {16 + 32 - 32} \cr & \Rightarrow \sqrt {16} \cr & \Rightarrow 4 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = \frac{{\sqrt 3 + 1}}{2} \cr & {x^2} + \frac{1}{{{x^2}}} = {\left( {\frac{{\sqrt 3 + 1}}{2}} \right)^2} - 2 \cr & = \frac{{4 + 2\sqrt 3 }}{4} - 2 \cr & = \frac{{4 + 2\sqrt 3 - 8}}{4} \cr & = \frac{{ - 4 + 2\sqrt 3 }}{4} \cr & = \frac{{ - 2 + \sqrt 3 }}{2} \cr & {x^4} + \frac{1}{{{x^4}}} = {\left( {\frac{{ - 2 + \sqrt 3 }}{2}} \right)^2} - 2 \cr & = \frac{{4 + 3 - 4\sqrt 3 - 8}}{4} \cr & = \frac{{ - 1 - 4\sqrt 3 }}{4} \cr} $$

x = 2 + √3
x² = 7 + 4√3
x⁴ - 8x³ + 16x²
= x²(x² - 8x + 16)
= x²(x - 4)²
= (7 + 4√3)(2 + √3 - 4)²
= (7 + 4√3)(2 - √3)²
= (7 + 4√3)(7 - 4√3)
= 1

$$\eqalign{ & \frac{{7 + 8 \times 8 \div 8\,{\text{of }}8 + 8 \div 8 \times 4\,{\text{of }}4}}{{4 \div 4\,{\text{of }}4 + 4 \times 4 \div 4 - 4 \div 4\,{\text{of}}\,{\text{2}}}} \cr & = \frac{{7 + 8 \times 8 \div 64 + 8 \div 8 \times 16}}{{4 \div 16 + 4 \times 4 \div 4 - 4 \div 8}} \cr & = \frac{{7 + 1 + 16}}{{\frac{1}{4} + 4 - \frac{1}{2}}} \cr & = \frac{{24}}{{\frac{{1 + 16 - 2}}{4}}} \cr & = \frac{{24 \times 4}}{{15}} \cr & = \frac{{32}}{5} \cr & = 6.4 \cr} $$

$$\eqalign{ & {\text{Put }}\sqrt {\frac{{1 - a}}{a}} = x \cr & \Rightarrow 3x + 9 = 19 - \frac{3}{x} \cr & \Rightarrow 3x + \frac{3}{x} = 10 \cr & \Rightarrow x = 3,\,\frac{1}{3} \cr & {\text{Now, }}\sqrt {\frac{{1 - a}}{a}} = 3 \cr & \Rightarrow \frac{{1 - a}}{a} = 9 \cr & \Rightarrow 10a = 1 \cr & \Rightarrow a = \frac{1}{{10}} \cr & {\text{and }}\sqrt {\frac{{1 - a}}{a}} = \frac{1}{3} \cr & \Rightarrow \frac{{1 - a}}{a} = \frac{1}{9} \cr & \Rightarrow 10a = 9 \cr & \Rightarrow a = \frac{9}{{10}} \cr & \therefore \,a = \frac{1}{{10}}\,{\text{and }}\frac{9}{{10}} \cr} $$