May 2026

If x³ + 1/x³ = 110, then x + 1/x = Question: If \[ x^3+\frac{1}{x^3}=110, \] then \[ x+\frac{1}{x}= \] (a) 5 (b) 10 (c) 15 (d) none of these Solution: Using identity: \[ \left(x+\frac{1}{x}\right)^3 = x^3+\frac{1}{x^3} + 3\left(x+\frac{1}{x}\right) \] Substituting the given value: \[ \left(x+\frac{1}{x}\right)^3 = 110+3\left(x+\frac{1}{x}\right) \] Let \[ x+\frac{1}{x}=a \] Then \[ […]

If x² + 1/x² = 102, then x − 1/x = Question: If \[ x^2+\frac{1}{x^2}=102, \] then \[ x-\frac{1}{x}= \] (a) 8 (b) 10 (c) 12 (d) 13 Solution: Using identity: \[ \left(x-\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}-2 \] Substituting the given value: \[ \left(x-\frac{1}{x}\right)^2 = 102-2 \] \[ \left(x-\frac{1}{x}\right)^2 = 100 \] \[ x-\frac{1}{x} = \sqrt{100} \]

If x + 1/x = 3, then x⁶ + 1/x⁶ = Question: If \[ x+\frac{1}{x}=3, \] then \[ x^6+\frac{1}{x^6}= \] (a) 927 (b) 414 (c) 364 (d) 322 Solution: First find \[ x^2+\frac{1}{x^2} \] Using identity: \[ \left(x+\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}+2 \] \[ (3)^2 = x^2+\frac{1}{x^2}+2 \] \[ 9 = x^2+\frac{1}{x^2}+2 \] \[ x^2+\frac{1}{x^2} = 7

If x + 1/x = 4, then x⁴ + 1/x⁴ = Question: If \[ x+\frac{1}{x}=4, \] then \[ x^4+\frac{1}{x^4}= \] (a) 196 (b) 194 (c) 192 (d) 190 Solution: First find \[ x^2+\frac{1}{x^2} \] Using identity: \[ \left(x+\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}+2 \] Substituting the given value: \[ (4)^2 = x^2+\frac{1}{x^2}+2 \] \[ 16 = x^2+\frac{1}{x^2}+2 \]

If x + 1/x = 2, then x³ + 1/x³ = Question: If \[ x+\frac{1}{x}=2, \] then \[ x^3+\frac{1}{x^3}= \] (a) 64 (b) 14 (c) 8 (d) 2 Solution: Using identity: \[ \left(x+\frac{1}{x}\right)^3 = x^3+\frac{1}{x^3} + 3\left(x+\frac{1}{x}\right) \] Substituting the given value: \[ (2)^3 = x^3+\frac{1}{x^3}+3(2) \] \[ 8 = x^3+\frac{1}{x^3}+6 \] \[ x^3+\frac{1}{x^3} =

If x + 1/x = 5, then x² + 1/x² = Question: If \[ x+\frac{1}{x}=5, \] then \[ x^2+\frac{1}{x^2}= \] (a) 25 (b) 10 (c) 23 (d) 27 Solution: Using identity: \[ \left(x+\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}+2 \] Substituting the given value: \[ (5)^2 = x^2+\frac{1}{x^2}+2 \] \[ 25 = x^2+\frac{1}{x^2}+2 \] \[ x^2+\frac{1}{x^2} = 25-2 \]

Find xy/(x² + y²) Question: If \[ x^2+y^2-xy=3 \] and \[ y-x=1 \] find: \[ \frac{xy}{x^2+y^2} \] Solution: Using identity: \[ (y-x)^2=x^2+y^2-2xy \] Substituting the given value: \[ 1^2=x^2+y^2-2xy \] \[ 1=x^2+y^2-2xy \] Given: \[ x^2+y^2-xy=3 \] Subtracting the two equations: \[ (x^2+y^2-xy)-(x^2+y^2-2xy)=3-1 \] \[ xy=2 \] Now, \[ x^2+y^2-xy=3 \] \[ x^2+y^2-2=3 \] \[

Find the Required Value Question: If \[ \frac ab+\frac ba=2 \] find: \[ \left(\frac ab\right)^{100} – \left(\frac ba\right)^{100} \] Solution: Using identity: \[ \left(x-y\right)^2 = \left(x+y\right)^2-4xy \] Let \[ x=\frac ab,\qquad y=\frac ba \] Then, \[ xy=\frac ab\cdot\frac ba=1 \] Given: \[ x+y=2 \] Therefore, \[ (x-y)^2=(2)^2-4(1) \] \[ =4-4 \] \[ =0 \] \[

Find the Required Value Question: If \[ a+b+c=6 \] and \[ \frac1a+\frac1b+\frac1c=\frac32 \] find: \[ \frac ab+\frac ac+\frac ba+\frac bc+\frac ca+\frac cb \] Solution: Given: \[ \frac1a+\frac1b+\frac1c=\frac32 \] Taking LCM: \[ \frac{ab+bc+ca}{abc}=\frac32 \] Using identity: \[ (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \] Now, \[ \frac ab+\frac ac+\frac ba+\frac bc+\frac ca+\frac cb \] \[ = \frac{a^2+b^2+c^2}{ab+bc+ca} \] Also,

Find ab²c² + a²bc² + a²b²c Question: If \[ \frac1a+\frac1b+\frac1c=1 \] and \[ abc=2 \] find: \[ ab^2c^2+a^2bc^2+a^2b^2c \] Solution: Given: \[ \frac1a+\frac1b+\frac1c=1 \] Taking LCM: \[ \frac{bc+ca+ab}{abc}=1 \] Since \[ abc=2 \] \[ \frac{ab+bc+ca}{2}=1 \] \[ ab+bc+ca=2 \] Now, \[ ab^2c^2+a^2bc^2+a^2b^2c \] \[ = abc(ab+bc+ca) \] Substituting the values: \[ = 2\times2 \] \[