Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & a = 2,{\text{ }}b = - 3 \cr & {\text{So,}}27{a^3} - 54{a^2}b + 36a{b^2} - 8{b^3} \cr} $$
$$ = 27 \times {\left( 2 \right)^3} - 54 \times {\left( 2 \right)^2} \times \left( { - 3} \right) + $$       $$36 \times $$ $$\left( 2 \right) \times $$ $${\left( { - 3} \right)^2}$$ $$ -\, 8 \times $$ $${\left( { - 3} \right)^3}$$
$$\eqalign{ & = 27 \times 8 - 54 \times 4 \times - 3 + 36 \times 2 \times 9 - 8 \times - 27 \cr & = 216 + 216 + 648 + 648 \cr & = 1728 \cr} $$

$$\eqalign{ & {\text{But }}a = 1 \cr & {a^3} + \frac{1}{{{a^3}}} = 2 \cr & \Rightarrow {1^3} + \frac{1}{{{1^3}}} = 2 \cr & \Rightarrow 2 = 2{\text{ }}\left( {{\text{Satisfy}}} \right) \cr & {\text{So, }}\frac{{{a^2} + 1}}{a} \cr & = \frac{{{1^2} + 1}}{1} \cr & = 2 \cr} $$

$$\eqalign{ & {\text{Let }}\frac{a}{{q - r}} = \frac{b}{{r - p}} = \frac{c}{{p - q}} = k \cr & \frac{a}{{q - r}} = k{\text{ }}\left( {{\text{On multiplying by }}p} \right) \cr & pa = k\left( {pq - pr} \right)\,......(i) \cr & {\text{In the same way we can write}} \cr & {\text{qb = k}}\left( {qr - qp} \right)\,........(ii) \cr & {\text{And }}rc = k\left( {rp - rp} \right)\,.....iii) \cr & {\text{On adding equation (i), (ii) and (iii) }} \cr & pa + qb + rc \cr & = k\left( {pq - pr + qr - qp + rp - rq} \right) \cr & = 0 \cr} $$

$$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 3\sqrt 3 \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr} $$

$$\eqalign{ & \frac{{{a^2} + {b^2}}}{{{c^2}}} = \frac{{{b^2} + {c^2}}}{{{a^2}}} = \frac{{{c^2} + {a^2}}}{{{b^2}}} = \frac{1}{k} \cr & {\text{Put }}a = b = c = 1 \cr & \Rightarrow \frac{{1 + 1}}{1} + \frac{{1 + 1}}{1} + \frac{{1 + 1}}{1} = \frac{1}{k} \cr & \Rightarrow 2 = 2 = 2 = \frac{1}{k} \cr & \Rightarrow k = \frac{1}{2} \cr} $$

$$\eqalign{ & 2x + \frac{2}{{9x}} = 4 \cr & {\text{Multiply by }}\frac{3}{2}{\text{ on both sides}} \cr & \Rightarrow 3x + \frac{1}{{3x}} = 6 \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\left( {3x + \frac{1}{{3x}}} \right)^3} = {6^3} \cr & \Rightarrow 27{x^3} + \frac{1}{{27{x^3}}} + 3 \times 3x \times \frac{1}{{3x}}\left( {3x + \frac{1}{{3x}}} \right) = 216 \cr & \Rightarrow 27{x^3} + \frac{1}{{27{x^3}}} + 3 \times 6 = 216 \cr & \Rightarrow 27{x^3} + \frac{1}{{27{x^3}}} = 216 - 18 \cr & \Rightarrow 27{x^3} + \frac{1}{{27{x^3}}} = 198 \cr} $$

$$\eqalign{ & {x^2} + \frac{{{\text{ }}1}}{{{x^2}}} = 2 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} + 2.x.\frac{1}{x} = 2 \cr & \Rightarrow x - \frac{1}{x} = 2 - 2 \cr & \Rightarrow x - \frac{1}{x} = 0 \cr} $$

$$\eqalign{ & pq\left( {p + q} \right) = 1 \cr & \Rightarrow p + q = \frac{1}{{pq}} \cr & \,\,\,\,\,\,\, \text{Cubing both side} \cr & \Rightarrow {p^3} + {q^3} +3 pq\left( {p + q} \right) = \frac{1}{{{p^3}{q^3}}} \cr & \,\,\,\,\,\,\, \text{Puting the value of } pq = \frac{1}{\left({p + q}\right)} \cr & \Rightarrow \frac{1}{{{p^3}{q^3}}} - {p^3} - {q^3} = \left( {\frac{{3}}{{p + q}}} \right)\left( {p + q} \right) \cr & \Rightarrow \frac{1}{{{p^3}{q^3}}} - {p^3} - {q^3} = 3 \cr} $$

$${p^3} - {q^3} = \left( {p - q} \right)\left\{ {{p^2} + {q^2} + pq} \right\}$$
$$ \Rightarrow \left( {p - q} \right)\left\{ {{{\left( {p + q} \right)}^2} - xpq} \right\} = $$       $$\left( {p - q} \right)$$ $$\left( {{p^2} + {q^2} + pq} \right)$$
$$\eqalign{ & \Rightarrow {p^2} + {q^2} + 2pq - xpq = {p^2} + {q^2} + pq \cr & \Rightarrow 2pq - pq = xpq \cr & \Rightarrow pq = xpq \cr & \Rightarrow x = 1 \cr} $$

$$\eqalign{ & x = {\left( {0.25} \right)^{\frac{1}{2}}} \cr & x = \root 2 \of {0.25} = 0.5{\text{ }} \cr & y = {\left( {0.4} \right)^2} \cr & y = 0.16 \cr & z = {\left( {0.216} \right)^{\frac{1}{2}}} \cr & z = \root 2 \of {0.216} = 0.464 \cr & \therefore x > z > y \cr} $$