If \$\dfrac{x}{y}+\dfrac{y}{x}=-1, (x, y \neq 0)\$, then value of \$x^3-y^3\$ is:

if \$\dfrac{x}{y} +\dfrac{y}{x}=-1\$simplifies to \$x^2+xy+y^2=0\$as \$(x^3-y^3)=(x-y)(x^2+xy+y^2)=0\$ using above result

If \$x+\dfrac{1}{x}=7\$, find \$x^3+\dfrac{1}{x^3}\$.

It is given that \$x+\dfrac { 1 }{ x } =7\$, by taking cubes on both sides we get:\$\left( x+\dfrac { 1 }{ x } \right) ^{ 3 }=7^{ 3 }\\ \Rightarrow \left( x \right) ^{ 3 }+\left( \dfrac { 1 }{ x } \right) ^{ 3 }+\left( 3\times x\times \dfrac { 1 }{ x } \right) \left( x+\dfrac { 1 }{ x } \right) =343\quad \quad (\because \quad (a+b)^{ 3 }=a^{ 3 }+b^{ 3 }+3ab(a+b))\\ \Rightarrow x^{ 3 }+\dfrac { 1 }{ x^{ 3 } } +3\left( 7 \right) =343\quad \quad \quad \quad \quad \quad \quad \quad \left( Given\quad x+\dfrac { 1 }{ x } =7 \right) \\ \Rightarrow x^{ 3 }+\dfrac { 1 }{ x^{ 3 } } +\left( 3\times 7 \right) =343\$\$\Rightarrow x^{ 3 }+\dfrac { 1 }{ x^{ 3 } } +21=343\\ \Rightarrow x^{ 3 }+\dfrac { 1 }{ x^{ 3 } } =343-21\\ \Rightarrow x^{ 3 }+\dfrac { 1 }{ x^{ 3 } } =322\$Hence, \$x^{ 3 }+\dfrac { 1 }{ x^{ 3 } } =322\$.

If \$x = 1 + \sqrt { 2 } ,\$ then find the value of \$\left( x - \dfrac { 1 } { x } \right) ^ { 3 }\$.

Given, \$\\x=1+\sqrt{2}\$.Now,\$\left(x-\dfrac{1}{x}\right)^3=\left(1+\sqrt{2}-(\dfrac{1}{1+\sqrt{2}})\right)^3\$\$\\=\left[1+\sqrt{2}-(\dfrac{1}{\sqrt{2}+1})\times(\dfrac{\sqrt{2}-1}{\sqrt{2}-1})\right]^3\$\$\\=\left[1+\sqrt{2}-(\dfrac{\sqrt{2}-1}{2-1})\right]^3\$\$\\=\left[1+\sqrt{2}-(\sqrt{2}-1)\right]^3\$\$\\=(1+\sqrt{2}-\sqrt{2}+1)^3=2^3=8\$.Therefore, option \$C\$ is correct.

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>>Class 9
>>Maths
>>Polynomials
>>Algebraic Identities
>> $Identities Involving the Sq...$

Identities Involving the Squares and Products of Binomials

Q: Solve \$(a+b+c)^{3}\$

\$(a+b+c)³=(a³+3a²b+3ab²+b³)+3(a²+2ab+b²)c+3(a+b)c2+c³\\ \$\$(a+b+c)³=a³+b³+c³+3a²b+3a²c+3ab²+3b²c+3ac²+3bc²+6abc\$\$(a+b+c)³=(a³+b³+c³)+(3a²b+3a²c+3abc)+(3ab²+3b²c+3abc)+(3ac²+3bc²+3abc)−3abc\$\$(a+b+c)³=(a³+b³+c³)+3a(ab+ac+bc)+3b(ab+bc+ac)+3c(ac+bc+ab)−3abc\$\$(a+b+c)³=(a³+b³+c³)+3(a+b+c)(ab+ac+bc)−3abc\$\$(a+b+c)³=(a³+b³+c³)+3[(a+b+c)(ab+ac+bc)−abc]\$

Q: If \$a + b + c = 0\$ then \$a^3 + b^3 + c^3\$ is

Since, \$a^3 + b^3 + c^3 - 3abc = (a + b + c) ( a^2 + b^2 + c^2 -bc- ca - ab)\$Given, \$a + b + c = 0\$\$\therefore a^3 + b^3 + c^3 - 3abc = 0 \$\$\therefore a^3 + b^3 + c^3 = 3abc \$Option B is correct.

Q: Factorise : \$(a - b)^3 + (b - c)^3 + (c - a)^3\$.

We know, \$(a-b)^3=a^3-b^3-3a^2b+3ab^2\$\$\implies\$ \$(a-b)^3=a^3-b^3-3ab(a-b)\$.Then,\$(a-b)^3+(b-c)^3+(c-a)^3\$\$=(a^3-3a^2b+3ab^2-b^3)+(b^3-3b^2c+3bc^2-c^3)+(c^3-3c^2a+3ca^2-a^3)\$\$=-3a^2b+3ab^2-3b^2c+3bc^2-3c^2a+3ca^2\$\$=3a^2(c-b)+3b^2(a-c)+3c^2(b-a)\$.Therefore, option \$C\$ is correct.

Q: If \$2x+3y=12\$ and \$xy=6\$, find the value of \$8x^{3}+27y^{3}\$.

\$2x+3y=12\$ \$xy=6\$\$8x^{3}+27y^{3}\$\$=(2x)^{3}+(3y)^{3}\$\$=a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-ab)\$\$8x^{3}+27y^{3}=(2x+3y)((2x)^{2}+(3y)^{2}-(2x)(3y))\$\$=(12)[(2+3y)^{2}-12 xy- 6 xy]\$\$=(12)[(12)^{2}-18(6)]\$\$= 12[144-108]\$\$= 12\times 36\$\$= 432\$.

Q: Verify \$x^{3} - y^{3} = (x - y)(x^{2} + xy + y^{2})\$ using some non-zero positive integers and check by actual multiplication. Can you call theses as identities?

To prove: \$x^3-y^3=(x-y)(x^2+xy+y^2)\$Consider the right hand side (\$RHS\$) and expand it as follows:\$(x-y)(x^2+xy+y^2)=x^3+x^2y+xy^2-yx^2-xy^2-y^3\\=(x^3-y^3)+(x^2y+xy^2+x^2y-xy^2)=x^3-y^3=LHS\$ Hence proved.Yes, we can call it as an identity: For example:Let us take \$x=2\$ and \$y=1\$ in \$x^3-y^3=(x-y)(x^2+xy+y^2)\$ then the \$LHS\$ and \$RHS\$ will be equal as shown below:\$2^3-1^3=7\$ and \$(2-1)(2^2+(2\times 1)+1^2)=1(5+2)=1\times 7=7\$Therefore, \$LHS=RHS\$Hence, \$x^3-y^3=(x-y)(x^2+xy+y^2)\$ can be used as an identity.

Q: The sum of two number is 25 and their difference is 13. Find their product.

Let two number be \$x\$ and \$y\$ \$x+y=25\$ …… (1) \$x-y=13\$ …….. (2) On solving equations (1) and (2), we get \$ 2x=38 \$ \$ x=19 \$ \$ y=6 \$ Since, the product of these two numbers \$ =19\times 6 \$ \$ =114 \$ Hence, this is the answer.

Q: if \$3x+2y=12\$ and \$xy=6\$ then find the value of \$ { 27x }^{ 3 }+{ 8y }^{ 3 }\$

It is given that \$3x+2y=12\$ and \$xy=6\$.We find the value of \$27x^3+8y^3\$ as shown below:\$27x^3+8y^3\\ =(3x)^3+(2y)^3\$\$ =(3x+2y)[(3x)^2-(3x\times 2y)+(2y)^2]\quad (\because \quad a³+b³=(a+b)(a²-ab+b²))\$\$ =12[(3x)^2-(6\times xy)+(2y)^2]\quad \quad \quad (\because \quad 3x+2y=12,\quad xy=6)\$\$ =12[(3x)^2-(6\times 6)+(2y)^2]\$\$ =12[(3x)^2-36+(2y)^2]\$\$ =12[(3x)^2+(2y)^2+(2\times 3x\times 2y)-(2\times 3x\times 2y)-36]\$\$ =12[(3x+2y)^2-12xy-36]\$\$ =12[(12)^2-(12\times 6)-36]\$\$ =12(144-72-36)\$\$ =12(144-108)\$\$ =12\times 36\$\$ =432\$Hence, \$27x^3+8y^3=432\$.