May 2026

Value of {(2.3)³ − 0.027}/{(2.3)² + 0.69 + 0.09} Value of {(2.3)³ − 0.027}/{(2.3)² + 0.69 + 0.09} The value of \[ \frac{(2.3)^3-0.027}{(2.3)^2+0.69+0.09} \] is (a) \(2\) (b) \(3\) (c) \(2.327\) (d) \(2.273\) Solution \[ \frac{(2.3)^3-(0.3)^3}{(2.3)^2+(2.3)(0.3)+(0.3)^2} \] Using identity: \[ \frac{a^3-b^3}{a^2+ab+b^2}=a-b \] \[ =2.3-0.3 \] \[ =2 \] Therefore, \[ \boxed{(a)\ 2} \] Next Question […]

Factorization of (a − b)³ + (b − c)³ + (c − a)³ Factorization of (a − b)³ + (b − c)³ + (c − a)³ The expression \[ (a-b)^3+(b-c)^3+(c-a)^3 \] can be factorized as (a) \((a-b)(b-c)(c-a)\) (b) \(3(a-b)(b-c)(c-a)\) (c) \(-3(a-b)(b-c)(c-a)\) (d) \((a+b+c)(a^2+b^2+c^2-ab-bc-ca)\) Solution Let \[ x=a-b,\quad y=b-c,\quad z=c-a \] Then, \[ x+y+z=(a-b)+(b-c)+(c-a)=0 \] Using

Factorization of (x + y)³ − (x − y)³ Factorization of (x + y)³ − (x − y)³ \[ (x+y)^3-(x-y)^3 \] can be factorized as (a) \(2y(3x^2+y^2)\) (b) \(2x(3x^2+y^2)\) (c) \(2y(3y^2+x^2)\) (d) \(2x(x^2+3y^2)\) Solution \[ (x+y)^3=x^3+3x^2y+3xy^2+y^3 \] \[ (x-y)^3=x^3-3x^2y+3xy^2-y^3 \] \[ (x+y)^3-(x-y)^3 \] \[ =x^3+3x^2y+3xy^2+y^3 -x^3+3x^2y-3xy^2+y^3 \] \[ =6x^2y+2y^3 \] \[ =2y(3x^2+y^2) \] Therefore, \[

Factors of 8a³ + b³ − 6ab + 1 Factors of 8a³ + b³ − 6ab + 1 The factors of \[ 8a^3+b^3-6ab+1 \] are (a) \((2a+b-1)(4a^2+b^2+1-3ab-2a)\) (b) \((2a-b+1)(4a^2+b^2-4ab+1-2a+b)\) (c) \((2a+b+1)(4a^2+b^2+1-2ab-b-2a)\) (d) \((2a-1+b)(4a^2+1-4a-b-2ab)\) Solution \[ (2a)^3+b^3+1^3-3(2a)(b)(1) \] Using identity: \[ x^3+y^3+z^3-3xyz =(x+y+z)(x^2+y^2+z^2-xy-yz-zx) \] \[ =(2a+b+1)(4a^2+b^2+1-2ab-b-2a) \] Therefore, \[ \boxed{(c)\ (2a+b+1)(4a^2+b^2+1-2ab-b-2a)} \] Next Question / Full

Factors of x³ − 1 + y³ + 3xy Factors of x³ − 1 + y³ + 3xy The factors of \[ x^3-1+y^3+3xy \] are (a) \((x-1+y)(x^2+1+y^2+x+y-xy)\) (b) \((x+y+1)(x^2+y^2+1-xy-x-y)\) (c) \((x-1+y)(x^2-1-y^2+x+y+xy)\) (d) \(3(x+y-1)(x^2+y^2-1)\) Solution \[ x^3+y^3+(-1)^3-3(x)(y)(-1) \] Using identity: \[ a^3+b^3+c^3-3abc =(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] \[ = (x+y-1)(x^2+y^2+1-xy+x+y) \] Therefore, \[ \boxed{(a)\ (x+y-1)(x^2+y^2+1-xy+x+y)} \] Next Question /

Factors of x³ − x²y − xy² + y³ Factors of x³ − x²y − xy² + y³ The factors of \[ x^3-x^2y-xy^2+y^3 \] are (a) \((x+y)(x^2-xy+y^2)\) (b) \((x+y)(x^2+xy+y^2)\) (c) \((x+y)^2(x-y)\) (d) \((x-y)^2(x+y)\) Solution \[ x^3-x^2y-xy^2+y^3 \] \[ =x^2(x-y)-y^2(x-y) \] \[ =(x-y)(x^2-y^2) \] \[ =(x-y)^2(x+y) \] \[ \boxed{(d)\ (x-y)^2(x+y)} \] Next Question / Full Exercise

Factorization of a³ + b³ + 3ab − 1 Factorization of a³ + b³ + 3ab − 1 The factorized form of \[ a^3+b^3+3ab-1 \] is _________ Solution \[ a^3+b^3+3ab-1 \] \[ =a^3+b^3+(-1)^3-3ab(-1) \] Using identity: \[ x^3+y^3+z^3-3xyz =(x+y+z)(x^2+y^2+z^2-xy-yz-zx) \] \[ = (a+b-1) \left(a^2+b^2+1-ab+b+a\right) \] \[ \boxed{(a+b-1)(a^2+b^2+1-ab+a+b)} \] Next Question / Full Exercise

Factorization of 1/xyz(x² + y² + z²) + 2(1/x + 1/y + 1/z) Factorization of 1/xyz(x² + y² + z²) + 2(1/x + 1/y + 1/z) The factorized form of \[ \frac{1}{xyz}(x^2+y^2+z^2)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \] is ___________ Solution \[ \frac{x^2+y^2+z^2}{xyz} + 2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \] \[ = \frac{x^2+y^2+z^2+2xy+2yz+2zx}{xyz} \] \[ = \frac{(x+y+z)^2}{xyz} \] \[ \boxed{\frac{(x+y+z)^2}{xyz}} \] Next Question /

If (3x − y/5) = 10 and xy = 5, Find 27x³ − y³/125 If (3x − y/5) = 10 and xy = 5, Find 27x³ − y³/125 If \[ 3x-\frac{y}{5}=10 \] and \[ xy=5, \] then the value of \[ 27x^3-\frac{y^3}{125} \] is ____________ Solution \[ \left(3x-\frac{y}{5}\right)^3 = (3x)^3-\left(\frac{y}{5}\right)^3 -3(3x)\left(\frac{y}{5}\right)\left(3x-\frac{y}{5}\right) \] \[ 10^3 =

Factorization of a⁴ + b⁴ − a²b² Factorization of a⁴ + b⁴ − a²b² The factorized form of \[ a^4+b^4-a^2b^2 \] is __________ Solution \[ a^4+b^4-a^2b^2 \] \[ =(a^2+b^2)^2-3a^2b^2 \] \[ =(a^2+b^2)^2-(\sqrt{3}ab)^2 \] \[ =(a^2+b^2-\sqrt{3}ab)(a^2+b^2+\sqrt{3}ab) \] \[ \boxed{(a^2+b^2-\sqrt{3}ab)(a^2+b^2+\sqrt{3}ab)} \] Next Question / Full Exercise