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: 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

Q: 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\$.

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: 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.

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\$.

Question 1

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

Accepted Answer

$(a+b+c)&#179;=(a&#179;+3a&#178;b+3ab&#178;+b&#179;)+3(a&#178;+2ab+b&#178;)c+3(a+b)c2+c&#179;\ $$(a+b+c)&#179;=a&#179;+b&#179;+c&#179;+3a&#178;b+3a&#178;c+3ab&#178;+3b&#178;c+3ac&#178;+3bc&#178;+6abc$$(a+b+c)&#179;=(a&#179;+b&#179;+c&#179;)+(3a&#178;b+3a&#178;c+3abc)+(3ab&#178;+3b&#178;c+3abc)+(3ac&#178;+3bc&#178;+3abc)&#8722;3abc$$(a+b+c)&#179;=(a&#179;+b&#179;+c&#179;)+3a(ab+ac+bc)+3b(ab+bc+ac)+3c(ac+bc+ab)&#8722;3abc$$(a+b+c)&#179;=(a&#179;+b&#179;+c&#179;)+3(a+b+c)(ab+ac+bc)&#8722;3abc$$(a+b+c)&#179;=(a&#179;+b&#179;+c&#179;)+3[(a+b+c)(ab+ac+bc)&#8722;abc]$

Question 2

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

Accepted Answer

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.

Question 3

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

Accepted Answer

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

Question 4

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

Accepted Answer

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, $x^{ 3 }+\dfrac { 1 }{ x^{ 3 } } =322$.

Question 5

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

Accepted Answer

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.

Question 6

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

Accepted Answer

$2x+3y=12$ $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$.

Question 7

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?

Accepted Answer

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.

Question 8

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

Accepted Answer

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.

Question 9

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

Accepted Answer

Let two number be $x$ and $y$&#10;&#10;$x+y=25$            &#8230;&#8230;&#10;(1)&#10;&#10;$x-y=13$            &#8230;&#8230;..&#10;(2)&#10;&#10;On solving equations (1) and (2), we get&#10;&#10;$ 2x=38 $&#10;&#10;$ x=19 $&#10;&#10;$ y=6 $&#10;&#10;Since, the product of these two numbers&#10;&#10;$ =19\times 6 $&#10;&#10;$ =114 $&#10;&#10;Hence, this is the answer.

Question 10

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

Accepted Answer

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$$ =(3x+2y)[(3x)^2-(3x\times 2y)+(2y)^2]\quad  (\because \quad a&#179;+b&#179;=(a+b)(a&#178;-ab+b&#178;))$$ =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, $27x^3+8y^3=432$.

Revise with Concepts

Identities Involving the Squares and Products of Binomials

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Introduction to Algebraic Identities

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Algebraic Identity of (x+y)³ and (x-y)³.

Algebraic Identities Of x³+y³ and x³-y³.

Algebraic Identity of x³+y³+z³-3xyz

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

If $a + b + c = 0$ then $a^{3} + b^{3} + c^{3}$ is

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

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

Factorise : $(a - b)^{3} + (b - c)^{3} + (c - a)^{3}$ .

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

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

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

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

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

Revise with Concepts

Identities Involving the Squares and Products of Binomials

More Complex Identities

Quick Summary With Stories

Introduction to Algebraic Identities

Square and Product of Binomials

Algebraic Identity of (x+y)³ and (x-y)³.

Algebraic Identities Of x³+y³ and x³-y³.

Algebraic Identity of x³+y³+z³-3xyz

Solve (a+b+c)3

If a+b+c=0 then a3+b3+c3 is

If yx​+xy​=−1,(x,y=0), then value of x3−y3 is:

If x+x1​=7, find x3+x31​.

Factorise : (a−b)3+(b−c)3+(c−a)3.

If 2x+3y=12 and xy=6, find the value of 8x3+27y3.

Verify x3−y3=(x−y)(x2+xy+y2) using some non-zero positive integers and check by actual multiplication. Can you call theses as identities?

If x=1+2​, then find the value of (x−x1​)3.

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

if 3x+2y=12 and xy=6 then find the value of 27x3+8y3

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

If $a + b + c = 0$ then $a^{3} + b^{3} + c^{3}$ is

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

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

Factorise : $(a - b)^{3} + (b - c)^{3} + (c - a)^{3}$ .

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

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

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

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