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

Factorization of x⁶ + 64y⁶ Factorization of x⁶ + 64y⁶ The polynomial \[ x^6+64y^6 \] on factorization gives ___________ Solution \[ x^6+64y^6 \] \[ =x^6+8x^3y^3+64y^6-8x^3y^3 \] \[ =(x^3+4y^3)^2-(2\sqrt{2}xy^{3/2})^2 \] \[ =(x^3-4x^2y+16xy^2-8y^3)(x^3+4x^2y+16xy^2+8y^3) \] \[ \boxed{(x^2-2xy+4y^2)(x^4+2x^3y+4x^2y^2+8xy^3+16y^4)} \] Next Question / Full Exercise

Factors of a + b + c + 2√ab − 2√bc − 2√ca Factors of a + b + c + 2√ab − 2√bc − 2√ca The factors of the expression \[ a+b+c+2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca} \] are ____________ Solution \[ a+b+c+2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca} \] \[ =(\sqrt{a}+\sqrt{b})^2-2\sqrt{c}(\sqrt{a}+\sqrt{b})+c \] \[ =(\sqrt{a}+\sqrt{b})^2 -2(\sqrt{a}+\sqrt{b})\sqrt{c} +(\sqrt{c})^2 \] \[ =(\sqrt{a}+\sqrt{b}-\sqrt{c})^2 \] \[ \boxed{(\sqrt{a}+\sqrt{b}-\sqrt{c})^2} \] Next

Factorization of x² + y² − z² − 2xy Factorization of x² + y² − z² − 2xy The polynomial \[ x^2+y^2-z^2-2xy \] on factorization gives _____________ Solution \[ x^2+y^2-z^2-2xy \] \[ =(x-y)^2-z^2 \] \[ =[(x-y)-z][(x-y)+z] \] \[ =(x-y-z)(x-y+z) \] \[ \boxed{(x-y-z)(x-y+z)} \] Next Question / Full Exercise

Factorization of 11x² − 10√3x − 3 Factorization of 11x² − 10√3x − 3 Factorization of the polynomial \[ 11x^2 – 10\sqrt{3}x – 3 \] gives __________ Solution \[ 11x^2 – 10\sqrt{3}x – 3 \] \[ =11x^2-11\sqrt{3}x+\sqrt{3}x-3 \] \[ =11x(x-\sqrt{3})+\sqrt{3}(x-\sqrt{3}) \] \[ =(x-\sqrt{3})(11x+\sqrt{3}) \] \[ \boxed{(x-\sqrt{3})(11x+\sqrt{3})} \] Next Question / Full Exercise

If 1/a + 1/b + 1/c = 1 and abc = 2, Find ab²c² + a²bc² + a²b²c If 1/a + 1/b + 1/c = 1 and abc = 2, Find ab²c² + a²bc² + a²b²c If \[ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 \] and \[ abc=2, \] then find \[ ab^2c^2+a^2bc^2+a^2b^2c. \] Solution Given: \[ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 \] Multiplying

Factorize a³ + (b − a)³ − b³ Factorize a³ + (b − a)³ − b³ The factorized form of \[ a^3+(b-a)^3-b^3 \] is Solution Let \[ x=a,\quad y=(b-a),\quad z=-b \] Then, \[ x+y+z = a+(b-a)-b \] \[ x+y+z=0 \] We know that if \(x+y+z=0\), then \[ x^3+y^3+z^3=3xyz \] Therefore, \[ a^3+(b-a)^3-b^3 = 3(a)(b-a)(-b) \]

Factorize x² − 1 − 2a − a² Factorize x² − 1 − 2a − a² Question: Factorize: \[ x^2 – 1 – 2a – a^2 \] Solution Rearranging the terms: \[ x^2 – 1 – 2a – a^2 = x^2-(a^2+2a+1) \] \[ = x^2-(a+1)^2 \] Using the identity \[ A^2-B^2=(A-B)(A+B) \] Therefore, \[ x^2-(a+1)^2

Factorize y² + (x − 1)y − x Factorize y² + (x − 1)y − x The factorized form of the expression \[ y^2+(x-1)y-x \] is Solution Given expression: \[ y^2+(x-1)y-x \] Split the middle term: \[ = y^2+xy-y-x \] Grouping the terms: \[ = y(y+x)-1(y+x) \] Taking common factor: \[ = (y-1)(y+x) \] Therefore,