Educational

If 9x² – b = (3x + 1/2)(3x – 1/2), then the value of b is Question: \[ 9x^2-b=(3x+\frac{1}{2})(3x-\frac{1}{2}) \] (a) 0 (b) \(\frac{1}{\sqrt{2}}\) (c) \(\frac{1}{4}\) (d) \(\frac{1}{2}\) Solution: \[ (3x+\frac{1}{2})(3x-\frac{1}{2}) \] \[ =(3x)^2-(\frac{1}{2})^2 \] \[ =9x^2-\frac{1}{4} \] \[ 9x^2-b=9x^2-\frac{1}{4} \] \[ b=\frac{1}{4} \] \[ \boxed{\frac{1}{4}} \] Next Question / Full Exercise

If 49a² – b = (7a + 1/2)(7a – 1/2), then the value of b is Question: \[ 49a^2-b=(7a+\frac{1}{2})(7a-\frac{1}{2}) \] (a) 0 (b) \(\frac{1}{4}\) (c) \(\frac{1}{\sqrt{2}}\) (d) \(\frac{1}{2}\) Solution: \[ (7a+\frac{1}{2})(7a-\frac{1}{2}) \] \[ =(7a)^2-(\frac{1}{2})^2 \] \[ =49a^2-\frac{1}{4} \] \[ 49a^2-b=49a^2-\frac{1}{4} \] \[ b=\frac{1}{4} \] \[ \boxed{\frac{1}{4}} \] Next Question / Full Exercise

If a/b + b/a = 1, then a² + b² = Question: \[ \frac{a}{b}+\frac{b}{a}=1, \] then \[ a^2+b^2= \] (a) 1 (b) -1 (c) 1/2 (d) 0 Solution: \[ \frac{a}{b}+\frac{b}{a}=1 \] \[ \frac{a^2+b^2}{ab}=1 \] \[ a^2+b^2=ab \] \[ \boxed{ab} \] Next Question / Full Exercise

If a/b + b/a = 1, then a² + b² = Question: \[ \frac{a}{b}+\frac{b}{a}=1, \] then \[ a^2+b^2= \] (a) 1 (b) -1 (c) 1/2 (d) 0 Solution: \[ \frac{a}{b}+\frac{b}{a}=1 \] \[ \frac{a^2+b^2}{ab}=1 \] \[ a^2+b^2=ab \] \[ \boxed{ab} \] Next Question / Full Exercise

The Product (x² – 1)(x⁴ + x² + 1) is Equal to Question: \[ (x^2-1)(x^4+x^2+1) \] is equal to (a) \(x^8-1\) (b) \(x^8+1\) (c) \(x^6-1\) (d) \(x^6+1\) Solution: \[ =(x^2-1)(x^4+x^2+1) \] \[ =(x^2)^3-1^3 \] \[ =x^6-1 \] \[ \boxed{x^6-1} \] Next Question / Full Exercise

The Product (a+b)(a-b)(a²-ab+b²)(a²+ab+b²) is Equal to Question: \[ (a+b)(a-b)(a^2-ab+b^2)(a^2+ab+b^2) \] is equal to (a) \[ a^6+b^6 \] (b) \[ a^6-b^6 \] (c) \[ a^3-b^3 \] (d) \[ a^3+b^3 \] Solution: \[ =(a^2-b^2)(a^2-ab+b^2)(a^2+ab+b^2) \] \[ =(a^2-b^2)(a^4+a^2b^2+b^4) \] \[ =a^6-b^6 \] \[ \boxed{a^6-b^6} \] Next Question / Full Exercise

Algebraic Identities Equation Form Solution Question: \[ \frac{ (a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3 }{ (a-b)^3+(b-c)^3+(c-a)^3 } \] Solution: \[ = \frac{ [(a-b)(a+b)]^3+[(b-c)(b+c)]^3+[(c-a)(c+a)]^3 }{ (a-b)^3+(b-c)^3+(c-a)^3 } \] \[ = \frac{ (a-b)^3(a+b)^3+(b-c)^3(b+c)^3+(c-a)^3(c+a)^3 }{ (a-b)^3+(b-c)^3+(c-a)^3 } \] \[ = \frac{ 3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a) }{ 3(a-b)(b-c)(c-a) } \] \[ = (a+b)(b+c)(c+a) \] \[ \boxed{(a+b)(b+c)(c+a)} \] Next Question / Full Exercise

If a + b + c = 9 and ab + bc + ca = 23, then a³ + b³ + c³ – 3abc = Question: If \[ a+b+c=9 \] and \[ ab+bc+ca=23, \] then \[ a^3+b^3+c^3-3abc= \] (a) 108 (b) 207 (c) 669 (d) 729 Solution: Using identity: \[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] Now,

If a^(1/3) + b^(1/3) + c^(1/3) = 0, then Question: If \[ a^{1/3}+b^{1/3}+c^{1/3}=0, \] then (a) \[ a+b+c=0 \] (b) \[ (a+b+c)^3=27abc \] (c) \[ a+b+c=3abc \] (d) \[ a^3+b^3+c^3=0 \] Solution: Let \[ x=a^{1/3}, \quad y=b^{1/3}, \quad z=c^{1/3} \] Then \[ x+y+z=0 \] Using identity: \[ x^3+y^3+z^3=3xyz \quad \text{when} \quad x+y+z=0 \] Substituting back:

If a + b + c = 0, then a²/bc + b²/ca + c²/ab = Question: If \[ a+b+c=0, \] then \[ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab} = \] (a) 0 (b) 1 (c) -1 (d) 3 Solution: Taking LCM: \[ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab} = \frac{a^3+b^3+c^3}{abc} \] Using identity: \[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] Given: \[ a+b+c=0 \] Therefore, \[ a^3+b^3+c^3-3abc=0