Educational

(a – b)³ + (b – c)³ + (c – a)³ = Question: \[ (a-b)^3+(b-c)^3+(c-a)^3= \] (a) \[ (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] (b) \[ (a-b)(b-c)(c-a) \] (c) \[ 3(a-b)(b-c)(c-a) \] (d) none of these Solution: Let \[ x=a-b \] \[ y=b-c \] \[ z=c-a \] Then \[ x+y+z=0 \] Using identity: \[ x^3+y^3+z^3=3xyz \quad \text{when} \quad x+y+z=0

If a + b + c = 9 and ab + bc + ca = 23, then a² + b² + c² = Question: If \[ a+b+c=9 \] and \[ ab+bc+ca=23, \] then \[ a^2+b^2+c^2= \] (a) 35 (b) 58 (c) 127 (d) none of these Solution: Using identity: \[ (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \] Substituting

If x³ – 1/x³ = 14, then x – 1/x = Question: If \[ x^3-\frac{1}{x^3}=14, \] then \[ x-\frac{1}{x}= \] (a) 5 (b) 4 (c) 3 (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: \[ \left(x-\frac{1}{x}\right)^3 = 14-3\left(x-\frac{1}{x}\right) \] Let \[ x-\frac{1}{x}=a \] Then \[ a^3=14-3a \]

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, \[ x^2+y^2-xy=3 \] \[ x^2+y^2-2=3 \] \[