Algebra — RRB NTPC Quantitative Aptitude

• Given: $x + \frac{1}{x} = 4$; value of $x^{4} + \frac{1}{x^{4}} = ?$

• Solution:
Step 1: Using the power-doubling formula, $x^{2} + \frac{1}{x^{2}} = 4^{2} - 2 = 16 - 2 = 14$
Step 2: Doubling the power again, $x^{4} + \frac{1}{x^{4}} = 14^{2} - 2 = 196 - 2 = 194$

• Exam Trick: Perfect-square +2 check: since $x^{4} + \frac{1}{x^{4}} = k^{2} - 2$, adding $2$ to the correct option must give a perfect square. Checking the correct option: $194 + 2 = 196 = 14^{2}$.

• Final Answer: 194 (Option 2)

• Given: $x^{4} + \frac{1}{x^{4}} = 727, x > 1$; value of $x - \frac{1}{x} = ?$

• Solution:
Step 1: Halving the power to positive sum, $x^{2} + \frac{1}{x^{2}} = \sqrt{727 + 2} = \sqrt{729} = 27$
Step 2: Halving the power to negative difference, $x - \frac{1}{x} = \sqrt{27 - 2} = \sqrt{25} = 5$

• Exam Trick: Direct nested radical halving: $\sqrt{\sqrt{727 + 2} - 2} = \sqrt{27 - 2} = 5$.

• Final Answer: 5 (Option 3)

• Given: $x - \frac{1}{x} = 5$; value of $x^{3} - \frac{1}{x^{3}} = ?$

• Solution:
Step 1: Using the cubic identity for difference, $x^{3} - \frac{1}{x^{3}} = a^{3} + 3a$
Step 2: Substituting $a = 5$, we get $5^{3} + 3(5) = 125 + 15 = 140$

• Exam Trick: Factor form $a(a^{2} + 3) = 5(25 + 3) = 5 imes 28 = 140$. Since the sign is negative, the answer must be greater than $5^{3} = 125$ and divisible by $28$, leaving only $140$.

• Final Answer: 140 (Option 1)

• Given: $2x - \frac{3}{x} = 6$; value of $8x^{3} - \frac{27}{x^{3}} = ?$

• Solution:
Step 1: Notice that $(2x)^{3} = 8x^{3}$ and \left(\frac{3}{x} ight)^{3} = \frac{27}{x^{3}}
Step 2: Using $A^{3} - B^{3} = (A - B)^{3} + 3AB(A - B)$ where $A = 2x, B = \frac{3}{x}$, and $AB = 6$
Step 3: $8x^{3} - \frac{27}{x^{3}} = 6^{3} + 3(6)(6) = 216 + 108 = 324$

• Exam Trick: Weighted cubic formula $k^{3} + 3pqk$: here $k = 6, p = 2, q = 3$. Since $k = 6$, both $6^{3}$ and $18(6)$ are multiples of $36$ and $108$: $216 + 108 = 324$ (avoid the common trap $6^{3} + 3(6) = 234$ which forgets the coefficients $2 imes 3 = 6$).

• Final Answer: 324 (Option 4)

• Given: $x^{2} + \frac{1}{x^{2}} = 7$ and $0 < x < 1$; value of $x^{4} - \frac{1}{x^{4}} = ?$

• Solution:
Step 1: Factoring difference of squares, x^{4} - \frac{1}{x^{4}} = \left(x^{2} + \frac{1}{x^{2}} ight)\left(x^{2} - \frac{1}{x^{2}} ight)
Step 2: Since $0 < x < 1$, $x^{2} < \frac{1}{x^{2}}$, so $x^{2} - \frac{1}{x^{2}}$ is negative: $x^{2} - \frac{1}{x^{2}} = -\sqrt{7^{2} - 4} = -\sqrt{45} = -3\sqrt{5}$
Step 3: Multiplying both parts, $x^{4} - \frac{1}{x^{4}} = 7 imes (-3\sqrt{5}) = -21\sqrt{5}$

• Exam Trick: For $0 < x < 1$, $x^{n} - \frac{1}{x^{n}}$ is always negative (eliminates Options 1 and 4 immediately). Also the coefficient must be a multiple of $7$, which leaves only $-21\sqrt{5}$.

• Final Answer: $-21\sqrt{5}$ (Option 2)

• Given: $x + \frac{1}{x} = 3, x > 1$; value of $x^{8} - \frac{1}{x^{8}} = ?$

• Solution:
Step 1: Break $x^{8} - \frac{1}{x^{8}}$ by halving powers: \left(x^{4} + \frac{1}{x^{4}} ight)\left(x^{2} + \frac{1}{x^{2}} ight)\left(x + \frac{1}{x} ight)\left(x - \frac{1}{x} ight)
Step 2: Find each factor: $x - \frac{1}{x} = \sqrt{3^{2} - 4} = \sqrt{5}$; $x^{2} + \frac{1}{x^{2}} = 3^{2} - 2 = 7$; $x^{4} + \frac{1}{x^{4}} = 7^{2} - 2 = 47$
Step 3: Multiply all four factors: $47 imes 7 imes 3 imes \sqrt{5} = 987\sqrt{5}$

• Exam Trick: Unit-digit & divisibility check: integer part is $47 imes 7 imes 3$. Unit digit $= 7 imes 7 imes 3 ightarrow 9 imes 3 = 7$, and it must be divisible by $3$ ($9+8+7=24$). Only $987\sqrt{5}$ ends in $7$ and is a multiple of $3$.

• Final Answer: $987\sqrt{5}$ (Option 3)

• Given: $x + \frac{1}{x} = 4$; value of $x^{5} + \frac{1}{x^{5}} = ?$

• Solution:
Step 1: Find power 2 and power 3 sums: $x^{2} + \frac{1}{x^{2}} = 4^{2} - 2 = 14$ and $x^{3} + \frac{1}{x^{3}} = 4^{3} - 3(4) = 52$
Step 2: Apply the 5th-power formula: x^{5} + \frac{1}{x^{5}} = \left(x^{2} + \frac{1}{x^{2}} ight)\left(x^{3} + \frac{1}{x^{3}} ight) - \left(x + \frac{1}{x} ight)
Step 3: $x^{5} + \frac{1}{x^{5}} = 14 imes 52 - 4 = 728 - 4 = 724$

• Exam Trick: Unit-digit & multiple rule: for any integer $x + \frac{1}{x} = k$, $x^{5} + \frac{1}{x^{5}}$ always has the same unit digit as $k$ (by Fermat's cycle $k^{5} \equiv k \pmod{10}$) and is divisible by $k$. Only $724$ ends in $4$ and is divisible by $4$.

• Final Answer: 724 (Option 1)

• Given: $x - \frac{1}{x} = 3$; value of $x^{7} - \frac{1}{x^{7}} = ?$

• Solution:
Step 1: Find power 2 and power 4 sums: $x^{2} + \frac{1}{x^{2}} = 3^{2} + 2 = 11 \Rightarrow x^{4} + \frac{1}{x^{4}} = 11^{2} - 2 = 119$
Step 2: Find power 3 difference: $x^{3} - \frac{1}{x^{3}} = 3^{3} + 3(3) = 36$
Step 3: Using x^{7} - \frac{1}{x^{7}} = \left(x^{4} + \frac{1}{x^{4}} ight)\left(x^{3} - \frac{1}{x^{3}} ight) + \left(x - \frac{1}{x} ight) = 119 imes 36 + 3 = 4284 + 3 = 4287

• Exam Trick: Unit-digit & remainder-modulo-9 check: $119 imes 36$ has unit digit $9 imes 6 ightarrow 4$, and adding $3$ gives unit digit $7$ (only 4287). Also $36$ is a multiple of $9$, so $119 imes 36 + 3$ leaves remainder $3$ when divided by $9$ ($4+2+8+7 = 21 \equiv 3 \pmod{9}$).

• Final Answer: 4287 (Option 4)

• Given: $x = 5 + 2\sqrt{6}$; value of $x^{2} + \frac{1}{x^{2}} = ?$

• Solution:
Step 1: Check the difference of squares of the surd terms: $5^{2} - (2\sqrt{6})^{2} = 25 - 24 = 1$
Step 2: Rationalizing gives the conjugate $\frac{1}{x} = 5 - 2\sqrt{6}$, so $x + \frac{1}{x} = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 10$
Step 3: $x^{2} + \frac{1}{x^{2}} = 10^{2} - 2 = 98$

• Exam Trick: When $a^{2} - b = 1$ in $x = a + \sqrt{b}$, directly use $x + \frac{1}{x} = 2a = 2(5) = 10$, so $x^{2} + \frac{1}{x^{2}} = (2a)^{2} - 2 = 100 - 2 = 98$.

• Final Answer: 98 (Option 2)

• Given: $x + \frac{1}{x + 3} = -1$; value of $(x + 3)^{7} + \frac{1}{(x + 3)^{7}} = ?$

• Solution:
Step 1: Add $3$ to both sides to match the denominator $(x + 3)$: $(x + 3) + \frac{1}{x + 3} = -1 + 3 = 2$
Step 2: When a term plus its reciprocal equals $2$, the term itself equals $1$: $x + 3 = 1$
Step 3: Substitute $x + 3 = 1$: $(1)^{7} + \frac{1}{(1)^{7}} = 1 + 1 = 2$

• Exam Trick: Adding $3$ to RHS gives $-1 + 3 = 2$, which is the standard $u + \frac{1}{u} = 2 \Rightarrow u = 1$ case; hence $u^{n} + \frac{1}{u^{n}} = 2$ for any power $n$.

• Final Answer: 2 (Option 3)