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, $27x^3+8y^3=432$.

Algebraic Identities - II

Language

Rate

REVISE WITH CONCEPTS

More Complex Identities

Geometric Verification of Cubic Identities

QUICK SUMMARY WITH STORIES

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 $a^{3} + b^{3} + c^{3}$ 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 $27 x^{3} + 8 y^{3}$

Algebraic Identities - II

Language

Rate

REVISE WITH CONCEPTS

More Complex Identities

Geometric Verification of Cubic Identities

QUICK SUMMARY WITH STORIES

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 $27 x^{3} + 8 y^{3}$