Algebra Expressions MCQ Questions

Solution: $\sqrt[3]{\mathrm{p}\left( \mathrm{p}^2+3\mathrm{p}+3 \right) +1}\,\,$

$=\,\,\sqrt[3]{\mathrm{p}^3+3\mathrm{p}^2.1+3\mathrm{p}.1^2+1}$

$=\,\,\sqrt[3]{\left( \mathrm{p}+1 \right) ^3}$

= p + 1 = 999 + 1 = 1000

Solution: y + $\frac{1}{\mathrm{z}}$ = 1 ⇒ y = 1 – $\frac{1}{\mathrm{z}}$= $\frac{\mathrm{z}-1}{\mathrm{z}}$

Now x + $\frac{1}{\mathrm{y}}$= 1

⇒ xy + 1 = y

⇒ xy + 1 = $\frac{\mathrm{z}-1}{\mathrm{z}}$

⇒ xyz + z = z –1

⇒ xyz = –1

Solution: x² + y² + 2x + 1 = 0

⇒ x² + 2x + 1 + y² = 0

⇒ (x + 1)² + y² = 0

⇒ x + 1 = 0 and y = 0

⇒ x – 1 and y = 0

∴ x³¹ + y³⁵ = – 1 + 0 = –1

Solution: x² + y² + 2x + 1 = 0

⇒ x² + 2x + 1 + y² = 0

⇒ (x + 1)² + y² = 0

⇒ x + 1 = 0 and y = 0

⇒ x – 1 and y = 0

∴ x²⁰²⁰ + y²⁰²⁰ =  1 + 0 = 1

Solution: $\frac{\mathrm{a}}{1-\mathrm{a}}+\frac{\mathrm{b}}{1-\mathrm{b}}+\frac{\mathrm{c}}{1-\mathrm{c}}=1$

⇒$\frac{\mathrm{a}}{1-\mathrm{a}}+1+\frac{\mathrm{b}}{1-\mathrm{b}}+1+\frac{\mathrm{c}}{1-\mathrm{c}}+1=4$

⇒$\frac{\mathrm{a}+1-\mathrm{a}}{1-\mathrm{a}}+\frac{\mathrm{b}+1-\mathrm{b}}{1-\mathrm{b}}+\frac{\mathrm{c}+1-\mathrm{c}}{1-\mathrm{c}}=4$

⇒$\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=1$

Solution: $\frac{1}{1+\mathrm{a}}=\frac{\mathrm{a}}{\mathrm{a}+\mathrm{a}^2}=\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}$

Similarly, $\frac{1}{1+\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}$

Similarly, $\frac{1}{1+\mathrm{c}}=\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}$

∴$\frac{1}{1+\mathrm{a}}+\frac{1}{1+\mathrm{b}}+\frac{1}{1+\mathrm{c}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}=1$

Solution: Given, a + b + c = 0 ⇒ a = –(b+c) ⇒ a² = (b+c)²

Now, $\frac{\mathrm{a}^2}{\mathrm{a}^2-\mathrm{bc}}+\frac{\mathrm{b}^2}{\mathrm{b}^2-+\mathrm{ca}}+\frac{\mathrm{c}^2}{\mathrm{c}^2-\mathrm{ab}}$

$=\,\,\frac{\left( \mathrm{b}+\mathrm{c} \right) ^2}{\left( \mathrm{b}+\mathrm{c} \right) ^2-\mathrm{bc}}+\frac{\mathrm{b}^2}{\mathrm{b}^2+\mathrm{c}\left( \mathrm{b}+\mathrm{c} \right)}+\frac{\mathrm{c}^2}{\mathrm{c}^2+\mathrm{b}\left( \mathrm{b}+\mathrm{c} \right)}$

$=\frac{\mathrm{b}^2+2\mathrm{bc}+\mathrm{c}^2}{\mathrm{b}^2+\mathrm{bc}+\mathrm{c}^2}+\frac{\mathrm{b}^2}{\mathrm{b}^2+\mathrm{c}^2+\mathrm{bc}}+\frac{\mathrm{c}^2}{\mathrm{b}^2+\mathrm{c}^2+\mathrm{bc}}$

$=\frac{2\left( \mathrm{b}^2+\mathrm{bc}+\mathrm{c}^2 \right)}{\mathrm{b}^2+\mathrm{bc}+\mathrm{c}^2}=2$

Solution: a + b + c = 3

⇒ a² + b² + c² +2ab +2bc +2ca = 9 (squaring both side)

⇒ ab + bc + ca = $\frac{3}{2}$

Now, $\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}=1$

⇒ $\frac{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}{\mathrm{abc}}=1$

⇒ abc = ab + bc + ca = $\frac{3}{2}$

Solution: x² + y² + z² + 2 = 2(y – x)

⇒ x² + 2x + y² – 2y + z² + 2 = 0

⇒ (x² + 2x + 1) + (y² – 2y + 1) + z² = 0

⇒ (x + 1)² + (y – 1)² + z² = 0

⇒ x + 1 = 0 ⇒ x = – 1; y – 1 = 0 ⇒ y = 1; z = 0

∴ x³ + y³ + z³ = – 1 + 1 + 0 = 0

Solution: $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}=119$

⇒$\mathrm{x}^4+\frac{1}{\mathrm{x}^4}+2.\mathrm{x}^2.\frac{1}{\mathrm{x}^2}=121$

⇒ $\left( \mathrm{x}^2+\frac{1}{\mathrm{x}^2} \right) ^2=11^2$

⇒ $\left( \mathrm{x}^2+\frac{1}{\mathrm{x}^2} \right) =11$ ( x > 1)

⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}-2.\mathrm{x}.\frac{1}{\mathrm{x}}=9$

⇒$\left( \mathrm{x}-\frac{1}{\mathrm{x}} \right) ^2=3^2$

⇒$\mathrm{x}-\frac{1}{\mathrm{x}}=3$ ( X > 1)

Using Trick ⇒$\mathrm{x}^3-\frac{1}{\mathrm{x}^3}\,\,=\,\,3^3+3\times 3=36$

Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}$ = n³ – 3.n and If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}$ = n³ + 3.n