Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & {\text{We know that }} \cr & {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) \cr & {x^3} - p = \left( {x - 4} \right)\left( {{x^2} + 4x + 16} \right) \cr & \Rightarrow {x^3} - p = \left( {{x^3} - {4^3}} \right) \cr & \Rightarrow p = {4^3}{\text{ }}\left( {{\text{By comparison}}} \right) \cr & {\text{So, }}p = 64 \cr} $$

$$\eqalign{ & \frac{{5x}}{{4{x^2} + 10x + 1}} \cr & = \frac{5x}{{x\left( {4x + 10 + \frac{1}{x}} \right)}} \cr & = \frac{5}{{4x + \frac{1}{x} + 10}} \cr & = \frac{5}{{5 + 10}} \cr & = \frac{5}{{15}} \cr & = \frac{1}{3} \cr} $$

$$\eqalign{ & c + \frac{1}{c} = \sqrt 3 \cr & {\text{On cubing both side}} \cr & \Rightarrow {\left( {c + \frac{1}{c}} \right)^3} = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3.c.\frac{1}{c}\left( {c + \frac{1}{c}} \right) = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 0 \cr} $$

$${x^3} + {y^3} + {z^3} - 3xyz$$
$$ = \frac{1}{2}\left( {x + y + z} \right)$$  $$\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]$$
$$ = \frac{1}{2}\left( {222 + 223 + 225} \right)$$    $$\left[ {{{\left( {222 - 223} \right)}^2} + {{\left( {223 - 225} \right)}^2} + {{\left( {225 - 222} \right)}^2}} \right]$$
$$\eqalign{ & = \frac{1}{2}\left( {670} \right)\left( {1 + 4 + 9} \right) \cr & = \frac{1}{2} \times 670 \times 14 \cr & = 4690 \cr} $$

$$\eqalign{ & a + b + c = 0 \cr & {\text{Let, }}{a^3} + {b^3} + {c^3} = T \cr} $$
$$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \frac{1}{2}\left( {a + b + c} \right)$$      $$\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]$$
$$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( 0 \right)$$     $$\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]$$
$$\eqalign{ & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 0 \cr & \Rightarrow {a^3} + {b^3} + {c^3} = 3abc \cr} $$

$$\eqalign{ & \frac{1}{p} + \frac{1}{q} = \frac{1}{{p + q}} \cr & \Rightarrow \frac{{p + q}}{{pq}} = \frac{1}{{p + q}} \cr & \Rightarrow {\left( {p + q} \right)^2} = pq \cr & \Rightarrow \left( {{p^2} + {q^2} + 2pq - pq} \right) = 0 \cr & \Rightarrow \left( {{p^2} + {q^2} + pq} \right) = 0 \cr & {\text{Multiply by }}\left( {p - q} \right){\text{ both side}} \cr & \Rightarrow \left( {p - q} \right)\left( {{p^2} + {q^2} + pq} \right) = \left( {p - q} \right) \times 0 \cr & \Rightarrow {p^3} - {q^3} = 0 \cr} $$

$$\eqalign{ & \frac{{{{\left( {a + b} \right)}^2}}}{{{{\left( {a - b} \right)}^2}}} = \frac{{25}}{4} \cr & \Rightarrow \frac{{a + b}}{{a - b}} = \frac{5}{2} \cr & \Rightarrow {\text{By Componendo & Dividendo}} \cr & \Rightarrow \frac{{a + b + a - b}}{{a + b - a + b}} = \frac{{5 + 2}}{{5 - 2}} \cr & \Rightarrow \frac{{2a}}{{2b}} = \frac{7}{3} \cr & \Rightarrow \frac{a}{b} = \frac{7}{3} \cr & {\text{Now, the value of}} \cr & \Rightarrow {a^2} + {b^2} + 3ab \cr & \Rightarrow {7^2} + {3^2} + 3.7.3 \cr & \Rightarrow 49 + 9 + 63 \cr & \Rightarrow 121 \cr} $$

$$\eqalign{ & a - b = 2{\text{ }} \cr & {a^3} - {b^3} = 26 \cr & \Rightarrow {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) \cr & \Rightarrow 26 = \left( 2 \right)\left( {{a^2} + ab + {b^2}} \right) \cr & \Rightarrow 13 = \left( {{a^2} + ab + {b^2}} \right)\,....(i) \cr & \Rightarrow 4 = 13 + ab \cr & \Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab \cr & \Rightarrow {\left( 2 \right)^2} = {a^2} + {b^2} + ab - 3ab \cr & \Rightarrow 3ab = 9 \cr & \Rightarrow ab = 3 \cr & \therefore {\left( {a + b} \right)^2} \cr & = {\left( {a - b} \right)^2} + 4ab \cr & = 4 + 4 \times 3 \cr & = 16 \cr} $$

$$\eqalign{ & \mathop {\mathop {{{\left( {x - 4} \right)}^3}}\limits_ \Downarrow }\limits_{{a^3}} + \mathop {\mathop {{{\left( {y - 2} \right)}^3}}\limits_ \Downarrow }\limits_{{b^3}} + \mathop {\mathop {{{\left( {z - 3} \right)}^3}}\limits_ \Downarrow }\limits_{{c^3} - 3abc} - 3\left( {x - 4} \right)\left( {y - 2} \right)\left( {z - 3} \right) \cr & \Rightarrow a + b + c = x - 4 + y - 2 + z - 3 \cr & \Rightarrow a + b + c = x + y + z - 9 \cr & \Rightarrow a + b + c = 9 - 9 \cr & \Rightarrow a + b + c = 0 \cr} $$
$${\text{So, }}{\left( {x - 4} \right)^3} + {\left( {y - 2} \right)^3} + $$     $${\left( {z - 3} \right)^3} - $$   $$3\left( {x - 4} \right)$$  $$\left( {y - 2} \right)$$ $$\left( {z - 3} \right)$$
$$ \Rightarrow 0$$

$$\eqalign{ & {\text{ }}x + \frac{1}{{9x}} = 4 \cr & {\text{Multiply by 3 both side}} \cr & \Rightarrow {\text{3}}x + \frac{1}{{3x}} = 12 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {\text{9}}{x^2} + \frac{1}{{9{x^2}}} + 2 \times 3x \times \frac{1}{{3x}} = 144 \cr & \Rightarrow {\text{9}}{x^2} + \frac{1}{{9{x^2}}} + 2 = 144 \cr & \Rightarrow {\text{9}}{x^2} + \frac{1}{{9{x^2}}} = 142 \cr} $$