Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & {a^3} = 117 + {b^3}{\text{ , }}a = 3 + b \cr & {a^3} - {b^3}{\text{ = 117}}\,......{\text{(i)}} \cr & a - b = 3\,........(ii) \cr & {\text{Put }}a = 5,{\text{ }}b = 2 \cr & {\text{Both equation satisfy}} \cr & {\text{Now, }}a + b = 5 + 2 = 7 \cr} $$

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 98 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 100 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 100 \cr & \Rightarrow x + \frac{1}{x} = 10 \cr & {\text{Cubing}}\,{\text{both}}\,{\text{sides}} \cr & \Rightarrow {\text{ }}{x^3} + \frac{1}{{{x^3}}} = {10^3} - 3 \times 10 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 1000 - 3 \times 10 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 970 \cr} $$

$$\eqalign{ & {\text{Given , }}x = y + z \cr & \therefore x - y - z = 0 \cr & {\text{Then }}{x^3} - {y^3} - {z^3} = 3xyz \cr} $$

$$\eqalign{ & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr & = \frac{{1 - d + 1 - a + 1 - b + 1 - c}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - \left( {a + b + c + d} \right)}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - 4}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = 0 \cr} $$

$${x^5} - 12{x^4} + 12{x^3} - 12{x^2} + 12x - 1$$
$$ = {x^5} - 11{x^4} - {x^4} + 11{x^3} + {x^3} - 11{x^2} - $$       $$\,{x^2} + $$ $$11x \,+\, $$ $$x - $$ $$1$$
$${\text{Put }}x = 11$$
$$ = {11^5} - {11.11^4} - {11^4} + {11.11^3} + {11^3} - $$       $${11.11^2} -\, $$ $$\,{11^2} \,+\, $$ $$11.11 +\, $$ $$11 - $$ $$1$$
$$\eqalign{ & = 11 - 1 \cr & = 10 \cr} $$

$$\eqalign{ & a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a} \cr & {\text{Put , }}a = \frac{1}{2},b = 2,c = - 1 \cr & \Rightarrow \frac{1}{2} + \frac{1}{2} = 2 - 1 = - 1 + 2 \cr & \Rightarrow 1 = 1 = 1 \cr & abc = \frac{1}{2} \times 2 \times - 1 \cr & \boxed{abc = - 1} \cr & {\text{Again put}} \cr & a = - \frac{1}{2},b = - 2,c = 1 \cr & \Rightarrow a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a} \cr & \Rightarrow - \frac{1}{2} - \frac{1}{2} = - 2 + 1 = 1 - 2 \cr & \Rightarrow - 1 = - 1 = - 1 \cr & {\text{Equation satisfied}} \cr & \Rightarrow abc = - \frac{1}{2} \times - 2 \times 1 \cr & \boxed{abc = + 1} \cr & {\text{So, }}abc{\text{ can be }} - 1{\text{ and }} + 1 \cr} $$

$$\eqalign{ & {a^3} - 7a - 6 \cr & \left( {a + 1} \right)\left( {{a^2} - a - 6} \right) \cr & \left( {a + 1} \right)\left( {a + 2} \right)\left( {a - 3} \right) \cr & {\text{Now}},{\text{Sum of factors}} \cr & \left( {a + 1} \right) + a + 2 + a - 3 = 3a \cr} $$

$$\eqalign{ & {p^3} + {q^3} + {r^3} - 3pqr = 4 \cr & a = q + r \cr & b = r + p \cr & c = p + q \cr & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc \cr & = {\left( {q + r} \right)^3} + {\left( {r + p} \right)^3} + {\left( {p + q} \right)^3} - 3\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right) \cr & = {q^3} + {r^3} + 3qr\left( {q + r} \right) + {r^3} + {p^3} + 3rp\left( {r + p} \right) + {p^3} + {q^3} + 3pq\left( {p + q} \right) - 3\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right) \cr & = 2{p^3} + 2{q^3} + 2{r^3} + 3pq\left( {p + q} \right) + 3qr\left( {q + r} \right) + 3rp\left( {r + p} \right) - 3\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right) \cr & = 2\left( {{p^3} + {q^3} + {r^3}} \right) + 3{p^2}q + 3p{q^2} + 3{p^2}r + 3q{r^2} + 3{r^2}p + 3r{p^2} - 3\left[ {\left( {p + q} \right)\left( {qr + qp + {r^2} + rp} \right)} \right] \cr & = 2\left( {4 + 3pqr} \right) + 3{p^2}q + 3p{q^2} + 3{q^2}r + 3q{r^2} + 3{r^2}p + 3r{p^2} - 3pqr - 3{p^2}q - 3p{r^2} - 3{p^2}r - 3{q^2}r - 3q{r^2} - 3{q^2}p - 3pqr \cr & = 8 + 6pqr - 6pqr \cr & = 8 \cr & \cr & {\bf{Alternate:}} \cr & r = q = 0 \cr & {p^3} + 0 + 0 - 0 = 4 \cr & {p^3} = 4,\,a = 0,\,b = p,\,c = p \cr & {a^3} + {b^3} + {c^3} - 3abc \cr & = 0 + {p^3} + {p^3} - 0 \cr & = 2{p^3} \cr & = 2 \times 4 \cr & = 8 \cr} $$

$$\eqalign{ & x = a + \frac{1}{a},\,y = a - \frac{1}{a} \cr & {\text{Let}} \Rightarrow a = 1 \cr & x = 1 + \frac{1}{1} = 2 \cr & y = 1 - 1 = 0 \cr & \sqrt {{x^4} + {y^4} - 2{x^2}{y^2}} \cr & = \sqrt {16 + 0 - 0} \cr & = 4 \cr} $$

$$\eqalign{ & {x^2} + {y^2} - xy = 5\frac{1}{4} \cr & {x^2} + {y^2} - xy = \frac{{21}}{4}........\left( {\text{i}} \right) \cr & {x^4} + {y^4} + {x^2}{y^2} = 17\frac{1}{{16}} \cr & {x^4} + {y^4} + {x^2}{y^2} = \frac{{273}}{{16}} \cr & {\left( {{x^2} + {y^2}} \right)^2} - {\left( {xy} \right)^2} = \frac{{273}}{{16}} \cr & \left( {{x^2} - xy + {y^2}} \right)\left( {{x^2} + xy + {y^2}} \right) = \frac{{273}}{{16}} \cr & \frac{{21}}{4}\left( {{x^2} + xy + {y^2}} \right) = \frac{{273}}{{16}} \cr & \left( {{x^2} + xy + {y^2}} \right) = \frac{{273}}{{16}} \times \frac{4}{{21}} \cr & \left( {{x^2} + xy + {y^2}} \right) = \frac{{13}}{4}........\left( {{\text{ii}}} \right) \cr & {\text{solve }}\left( {\text{i}} \right) + \left( {{\text{ii}}} \right) \cr & 2\left( {{x^2} + {y^2}} \right) = \frac{{21}}{4} + \frac{{13}}{4} \cr & 2\left( {{x^2} + {y^2}} \right) = \frac{{17}}{2} \cr & \left( {{x^2} + {y^2}} \right) = \frac{{17}}{4} \cr & {\text{solve }}\left( {\text{i}} \right) - \left( {{\text{ii}}} \right) \cr & - 2xy = \frac{{21}}{4} - \frac{{13}}{4} \cr & - 2xy = \frac{8}{4} \cr & - 2xy = 2 \cr & xy = - 1 \cr & {\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy \cr & {\left( {x - y} \right)^2} = \frac{{17}}{4} - 2 \times \left( { - 1} \right) \cr & {\left( {x - y} \right)^2} = \frac{{17}}{4} + 2 \cr & {\left( {x - y} \right)^2} = \frac{{25}}{4} \cr & x - y = \frac{5}{2} \cr} $$