Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

According to the question,
$$\because $$ x = z = 225
y = 226
⇒ x³ + y³ + z³ - 3xyz = ?
As we know,
x³ + y³ + z³ - 3xyz
= $$\frac{1}{2}$$ (x + y + z) [(x - y)² + (y - z)² + (z - x)²]
= $$\frac{1}{2}$$ [225 + 225 + 226] [(225 - 226)² + (226 - 225)² + (225 - 225)²]
= $$\frac{676}{2}$$ × [1 + 1 + 0]
= 676

$$\eqalign{ & {\text{According to the question,}} \cr & \Rightarrow {\left( {x + 3} \right)^3} + \frac{1}{{{{\left( {x + 3} \right)}^3}}} \cr & {\text{Let, }}t = x + 3,{\text{ }}x = t - 3 \cr & \Rightarrow {x^2} + x = 5{\text{, }}\left( {{\text{Given}}} \right) \cr & \Rightarrow \left( {x + 1} \right)x = 5 \cr & \Rightarrow \left( {t - 3 + 1} \right).\left( {t - 3} \right) = 5 \cr & \Rightarrow \left( {t - 2} \right).\left( {t - 3} \right) = 5 \cr & \Rightarrow {t^2} - 3t - 2t + 6 = 5 \cr & \Rightarrow {t^2} - 5t = - 1 \cr & \Rightarrow t + \frac{1}{t} = 5 \cr & \Rightarrow {\left( {x + 3} \right)^3} + \frac{1}{{{{\left( {x + 3} \right)}^3}}} \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} \cr & \Rightarrow \left( {t + \frac{1}{t}} \right)^3 - 3 \times t \times \frac{1}{t}\left( {t + \frac{1}{t}} \right) \cr & \Rightarrow {5^3} - 3 \times 5 \cr & \Rightarrow 125 - 15 \cr & \Rightarrow 110 \cr & \therefore {\left( {x + 3} \right)^3} + \frac{1}{{{{\left( {x + 3} \right)}^3}}} = 110 \cr} $$

$$\eqalign{ & {\text{Given }}m = - 4,n = - 2, \cr & {\text{Find }}{m^3} - 3{m^2} + 3m + 3n + 3{n^2} + {n^3} = ? \cr & {\text{Putting value of }}m{\text{ and }}n \cr & \Rightarrow {\left( { - 4} \right)^3} - 3{\left( { - 4} \right)^2} + 3\left( { - 4} \right) + 3\left( { - 2} \right) + 3{\left( { - 2} \right)^2} + {\left( { - 2} \right)^3} \cr & \Rightarrow - 64 - 48 - 12 - 6 + 12 - 8 \cr & \Rightarrow - 64 - 60 - 2 \cr & \Rightarrow - 126 \cr} $$

Here, x = 332, y = 333, z = 335
Find x³ + y³ + z³ - 3xyz
$$ = \frac{1}{2}\left( {x + y + z} \right)$$   $$\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]$$
$$ = \left( {\frac{{332 + 333 + 335}}{2}} \right)$$   $$\left[ {{{\left( {332 - 333} \right)}^2} + {{\left( {333 - 335} \right)}^2} + {{\left( {332 - 335} \right)}^2}} \right]$$
$$\eqalign{ & = \frac{{1000}}{2}\left[ {{1^2} + {2^2} + {3^2}} \right] \cr & = \frac{{1000}}{2}\left( {14} \right) \cr & = 7000{\text{ }} \cr} $$

$$\eqalign{ & {\text{If }}2 + x\sqrt 3 = \frac{1}{{2 + \sqrt 3 }} \cr & {\text{Find }}x = ? \cr & \Rightarrow 2 + x\sqrt 3 = \frac{1}{{2 + \sqrt 3 }} \cr & \Rightarrow 2 + x \times \sqrt 3 = \frac{{2 - \sqrt 3 }}{1} \cr & \Rightarrow 2 + x\sqrt 3 = 2 - \sqrt 3 \cr & \Rightarrow x = - 1 \cr} $$

$$\eqalign{ & \frac{{m - {a^2}}}{{{b^2} + {c^2}}} + \frac{{m - {b^2}}}{{{c^2} + {a^2}}} + \frac{{m - {c^2}}}{{{a^2} + {b^2}}} = 3 \cr & \Rightarrow m = ? \cr & \Rightarrow \frac{{m - {a^2}}}{{{b^2} + {c^2}}} + \frac{{m - {b^2}}}{{{c^2} + {a^2}}} + \frac{{m - {c^2}}}{{{a^2} + {b^2}}} = 1 + 1 + 1 \cr & {\text{Put }}m = {a^2} + {b^2} + {c^2}{\text{ from option (B)}} \cr} $$
$${\text{L}}{\text{.H}}{\text{.S}}{\text{. = }}\frac{{{a^2} + {b^2} + {c^2} - {a^2}}}{{{b^2} + {c^2}}} + $$      $$\frac{{{a^2} + {b^2} + {c^2} - {b^2}}}{{{c^2} + {a^2}}} + $$    $$\frac{{{a^2} + {b^2} + {c^2} - {c^2}}}{{{a^2} + {b^2}}}$$
$$\eqalign{ & \Rightarrow \frac{{{b^2} + {c^2}}}{{{b^2} + {c^2}}} + \frac{{{a^2} + {c^2}}}{{{c^2} + {a^2}}} + \frac{{{a^2} + {b^2}}}{{{a^2} + {b^2}}} \cr & \Rightarrow 1 + 1 + 1 = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr & \Rightarrow m = {a^2} + {b^2} + {c^2} \cr} $$

$$\eqalign{ & {\text{Given, }}m - 5n = 2 \cr & {\text{Find, }}{m^3} - 125{n^3} - 30mn \cr & \Rightarrow m - 5n = 2 \cr & \left( {{\text{Cubing both sides}}} \right) \cr & \Rightarrow {\left( {m - 5n} \right)^3} = {2^3} \cr & \Rightarrow {m^3} - 125{n^3} - 3m \times 5n\left( {m - 5n} \right) = 8 \cr & \Rightarrow {m^3} - 125{n^3} - 15mn \times 2 = 8 \cr & \Rightarrow {m^3} - 125{n^3} - 30mn = 8 \cr} $$

$$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = 2\ , . . . . . . (i) \cr & {\text{The value of }}{x^{12}} - \frac{1}{{{x^{12}}}} =\, ? \cr & \Rightarrow {\text{If }}x = 1 \cr & \Rightarrow x + \frac{1}{x} = 2 \cr & \Rightarrow 1 + 1 = 2 \cr & {\text{Then, }}{x^{12}} - \frac{1}{{{x^{12}}}} \cr & \Rightarrow {1^{12}} - \frac{1}{{{1^{12}}}} \cr & \Rightarrow 1 - 1 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = 1 \cr & {\text{Find }}\frac{{{x^2} + 3x + 1}}{{{x^2} + 7x + 1}} \cr & {\text{From equation (i)}} \cr & \Rightarrow x + \frac{1}{x} = 1 \cr & \Rightarrow {x^2} + 1 = x \cr & \Rightarrow \frac{{\left( {{x^2} + 1} \right) + 3x}}{{\left( {{x^2} + 1} \right) + 7x}} \cr & \Rightarrow \frac{{x + 3x}}{{x + 7x}} \cr & \Rightarrow \frac{{4x}}{{8x}} \cr & \Rightarrow \frac{1}{2} \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 2 \cr & {\text{Find , }}{x^7} + \frac{1}{{{x^5}}} \cr & \,\,\,\,\,\,\,x + \frac{1}{x} = 2 \cr & \Rightarrow \mathop \downarrow \limits_1 {\text{ }}\,\,\,\,{\text{ }}\mathop \downarrow \limits_1 \,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {\text{Let }}x = 1 \cr & {\text{To put value in question,}} \cr & \Rightarrow {x^7} + \frac{1}{{{x^5}}} \cr & \Rightarrow {{\text{1}}^7} + \frac{1}{{{1^5}}} \cr & \Rightarrow 1 + 1 \cr & \Rightarrow 2 \cr} $$