The factors of 8a^3 + b^3 – 6ab + 1 are

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 […]

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The factors of x^3 – 1 – y^3 + 3xy are

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 /

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The factors of x^3 – x^2y – xy^2 + y^3 are

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

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The Factorized form of a^3 + b^3 + 3ab – 1 is _________

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

The Factorized form of a^3 + b^3 + 3ab – 1 is _________ Read More »

The Factorized form of 1/xyz(x^2 + y^2 + z^2) + 2(1/x + 1/y + 1/z) is ___________

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 /

The Factorized form of 1/xyz(x^2 + y^2 + z^2) + 2(1/x + 1/y + 1/z) is ___________ Read More »

If 3x – y/5 = 10 and xy = 5, then the value of 27x^3 – y^3/125 is______________

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 =

If 3x – y/5 = 10 and xy = 5, then the value of 27x^3 – y^3/125 is______________ Read More »

The Factorization form of a^4 + b^4 – a^2b^2 is __________

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

The Factorization form of a^4 + b^4 – a^2b^2 is __________ Read More »

The polynomial ,x^6 + 64y^6 on factorization gives ___________

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

The polynomial ,x^6 + 64y^6 on factorization gives ___________ Read More »

The factors of the expression a + b + c + 2√ab – 2√bc – 2√ca are ____________

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

The factors of the expression a + b + c + 2√ab – 2√bc – 2√ca are ____________ Read More »

The polynomial x^2 + y^2 – z^2 – 2xy on factorization gives _____________

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

The polynomial x^2 + y^2 – z^2 – 2xy on factorization gives _____________ Read More »