Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

a³ + b³ + c³ = 3abc
If a + b + c = 0
(5x + 1) + (x - 3) + (6x - 8) = 0
12x - 10 = 0
12x = 10
x = $$\frac{5}{6}$$

$$\eqalign{ & \frac{1}{4} + \frac{{\left[ {{{\left( {20.35} \right)}^2} - {{\left( {8.35} \right)}^2}} \right] \times 0.0175}}{{{{\left( {1.05} \right)}^2} + \left( {1.05} \right)\left( {27.65} \right)}} \cr & = \frac{1}{4} + \frac{{\left[ {\left( {20.35 + 8.35} \right)\left( {20.35 - 8.35} \right)} \right] \times 0.0175}}{{\left( {1.05} \right) + \left[ {1.05 + 27.65} \right]}} \cr & = \frac{1}{4} + \frac{{\left[ {\left( {28.70} \right)\left( {12} \right) \times 0.0175} \right]}}{{\left( {1.05} \right)\left[ {28.70} \right]}} \cr & = \frac{1}{4} + \frac{{12 \times 5}}{{300}} \cr & = \frac{1}{4} + \frac{1}{5} \cr & = \frac{9}{{20}} \cr} $$

$$\eqalign{ & {a^3} + {b^3} = 218\,\& \,a + b = 2 \cr & {\left( {a + b} \right)^3} = {\left( 2 \right)^3} \cr & {a^3} + {b^3} + 3\left( {a + b} \right)\left( {ab} \right) = 8 \cr & 218 + 3\left( 2 \right)\left( {ab} \right) = 8 \cr & ab = \frac{{8 - 218}}{6} \cr & ab = \frac{{ - 210}}{6} = - 35 \cr & \sqrt {1 - ab} = \sqrt {1 - \left( { - 35} \right)} \cr & \sqrt {1 - ab} = \sqrt {1 + \left( {35} \right)} \cr & \sqrt {1 - ab} = 6 \cr} $$

[8(x + y)³ - 27(x - y)³] ÷ (5y - x) = Ax² + Bxy + Cy²
Let, x = 1, y = 1
[8(1 + 1)³ - 27 × 0] ÷ (5 - 1) = (A + B + C)
$$\frac{{8 \times 8}}{4}$$  = A + B + C
16 = A + B + C

$$\eqalign{ & {x^3} - 4{x^2} + 19 = 6\left( {x - 1} \right)......\left( 1 \right)\left( {{\text{Given}}} \right) \cr & \& \,\left( {{x^2} + \frac{1}{{x - 4}}} \right) = ?\left( {{\text{To find}}} \right) \cr & \to \frac{{{x^2}\left( {x - 4} \right) + 1}}{{x - 4}}......\left( 2 \right) \cr & {\text{from equation}}\left( 1 \right) \cr & {x^3} - 4{x^2} + 19 = 6x - 6 \cr & {x^3} - 4{x^2} + 1 + 18 = 6x - 6 \cr & {x^3} - 4{x^2} + 1 = 6x - 6 - 18 \cr & {x^3} - 4{x^2} + 1 = 6x - 24 \cr & {\text{putting the value of }}\left( {{x^3} - 4{x^2} + 1} \right){\text{in equation }}\left( 2 \right) \cr & {\text{we get,}} = \frac{{6x - 24}}{{x - 4}} = \frac{{6\left( {x - 4} \right)}}{{\left( {x - 4} \right)}} = 6 \cr} $$

$$\eqalign{ & \left( {5\sqrt 5 {x^3} - 3\sqrt 3 {y^3}} \right) \div \left( {\sqrt 5 x - \sqrt 3 y} \right) = A{x^2} + B{y^2} + Cxy \cr & 3A - B - \sqrt {15} C = ? \cr & \frac{{{{\left( {\sqrt 5 x} \right)}^3} - {{\left( {\sqrt 3 x} \right)}^3}}}{{\sqrt 5 x - \sqrt 3 y}} = A{x^2} + B{y^2} + Cxy \cr & 5{x^2} + 3{y^2} + \sqrt {15} xy \cr & A = 5,\,B = 3,\,C = \sqrt {15} \cr & 3A - B - \sqrt {15} C \cr & = 3 \times 5 - 3 - \sqrt {15} \times \sqrt {15} \cr & = - 3 \cr} $$

x³ + 27y³ + 64z³ = 36xyz ......(i)
since ⇒ a + b + c = 0
If a³ + b³ + c³ = 3abc
similarly from equation (i)
x + 3y + 4z = 0

$$\eqalign{ & x + y + z = 0,\,\frac{{3{y^2} + {x^2} + {z^2}}}{{2{y^2} - xz}} = ? \cr & {\text{Put }}x = 1,\,y = 1,\,z = - 2 \cr & \Rightarrow \frac{{3{{\left( y \right)}^2} + {x^2} + {z^2}}}{{2{y^2} - xz}} \cr & \Rightarrow \frac{{3{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( { - 2} \right)}^2}}}{{2{{\left( 1 \right)}^2} - \left( {1 \times \left( { - 2} \right)} \right)}} \cr & \Rightarrow \frac{{4 + 4}}{{2 + 2}} \cr & \Rightarrow \frac{8}{4} \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = - 14 \cr & x - \frac{1}{x} = \sqrt {{{\left( { - 14} \right)}^2} - 4} \cr & = \sqrt {196 - 4} \cr & = \sqrt {192} \cr & = 8\sqrt 3 \cr & \therefore \,x < - 1 \cr & \left( {x + \frac{1}{x}} \right)\left( {x - \frac{1}{x}} \right) = - 14x\left( {8\sqrt 3 } \right) \cr & {x^2} - \frac{1}{{{x^2}}} = + 112\sqrt 3 \cr} $$