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Give possible expressions for the length and breadth of the rectangle whose area is given by 4a^{2}+4a-3

Length (2a+3)

Breadth (2a-1)

            Or

Length = (2a-1)

Breadth =(2a+3)

Solution:

Given : Area of rectangle is 4a2+4a-3    …..(1)

Factorize equation 1, we get

=4a2+6a-2a-3

=2a(2a+3)-1(2a+3)

=(2a+3)(2a-1)

We know that area of rectangle is length × breadth

Hence

Length (2a+3)

Breadth (2a-1)

            Or

Length = (2a-1)

Breadth =(2a+3)

 

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    Find the value of

(i) x^{3}+y^{3}-12xy-64 when x+y=-4

(ii) x^{3}-8y^{3}-36xy-216 when x=2y+6

 

(i) 0

Solution :-

Here x+y=-4

We know that

a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

When a + b + c = 0, we get a^{3}+b^{3}+c^{3}=3abc           

Here,

x+y+4=0

So using the above identity, we get:

x^{3}+y^{3}+4^{3}=3 \times x \times y\times 4= 12xy \cdots \cdots (i)

Now

\\x^{3}+y^{3}-12xy+64=x^{3}+y^{3}+(4)^{3}-12xy\\ 12xy-12xy=0{from equation i}

Hence the answer is 0

(ii) 0

Solution :-

We know that

a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

When a + b + c = 0, we get a^{3}+b^{3}+c^{3}=3abc           

Here,

x-2y-6=0

So using the above identity, we get:

x^{3}+(-2y)^{3}-(-6)^{3}=3x(-2y)(-6)= 36xy

x^{3}+8y^{3}-216=36xy                                           ....................(i)

Now

\\x^{3}-8y^{3}-216-36xy=36xy-36xy =0

Hence the answer is 0.

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Without finding the cubes, factorize  (x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}

3(x-2y)(2y-3z)(3z-x)

solution :

given  (x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}

We know that

a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

When a + b + c = 0, we get a^{3}+b^{3}+c^{3}=3abc

Here,

(x-2y)+(2y-3z)+(3z-x)=0

So using the above identity, we get:

(x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}

=3(x-2y)(2y-3z)(3z-x)

Hence the answer is 3(x-2y)(2y-3z)(3z-x).

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Without actually calculating the cubes, find the value of

(i) \left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}-\left ( \frac{5}{6} \right )^{3}

(ii) (0.2)^{3}-(0.3)^{3}+(0.1)^{3}

(i) \frac{-5}{12}

Given \left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}-\left ( \frac{5}{6} \right )^{3}

We know that

a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

When a + b + c = 0, we get a^{3}+b^{3}+c^{3}=3abc

Here,

\frac{1}{2} +\frac{1}{3}- \frac{5}{6} =\frac{3+2-5}{6}=\frac{0}{6}=0

So using the above identity, we get:

\therefore \left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}+\left ( \frac{-5}{6} \right )^{3}=3 \times \frac{1}{2} \times \frac{1}{3}\times \frac{-5}{6}=\frac{-5}{12}

Hence the answer is \frac{-5}{12}

(ii)-0.018

Solution

Given: (0.2)^{3}-(0.3)^{3}+(0.1)^{3}

We know that

a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

When a + b + c = 0, we get a^{3}+b^{3}+c^{3}=3abc

Here,

0.2+(-0.3)+0.1=0

So,                              
(0.2)^{3}-(0.3)^{3}+(0.1)^{3}=3(0.2)(-0.3)(0.1)=-0.018

Hence the answer is -0.018.

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Factorize

(i) a^{3}-8b^{3}-64c^{3}-24abc

(ii) 2\sqrt{2}a^{3}+8b^{3}-27c^{3}+18\sqrt{2}abc

(i) (a-2b-4c)(a^{2}+4b^{2}+16c^{2}+2ab-8bc+4ac)

Solution:

Given: a^{3}-8b^{3}-64c^{3}-24abc

This can be written as:

a^{3}+(-2b)^{3}+(-4c)^{3}-3(a)(-2b)(-4c)

We know that

a^{3}+b^{3}+c^{3}-3abc=\left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2}-ab-bc-ac \right )

Using this identity we get

a^{3}+(-2b)^{3}+(-4c)^{3}-3(a)(-2b)(-4c)

=\left ( a+(-2b)+(-4c)\right )\left ( a^{2}-(-2b)^{2}+(-4c)^{2}-(a)(-2b)-(-2b)(-4c)-(a)(-4c) \right )\\     

=(a-2b-4c)(a^{2}+4b^{2}+16c^{2}+2ab-8bc+4ac)

         

(i) (\sqrt{2}a+2b-3c)(2a^{2}+4b^{2}+9c^{2}-2\sqrt{2}ab+6bc+3\sqrt{2}ca)

Solution:

Given: 2\sqrt{2}a^{3}+8b^{3}-27c^{3}+18\sqrt{2}abc

This can be written as:

(\sqrt{2}a)^{3}+(2b)^{3}-(3c)^{3}-3(\sqrt{2}a)\left (2b \right )(3c)

We know that

a^{3}+b^{3}+c^{3}-3abc=\left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2}-ab-bc-ac \right )

Using this identity we get

(\sqrt{2}a)^{3}+(2b)^{3}-(3c)^{3}-3(\sqrt{2}a)\left (2b \right )(3c)

=\left ((\sqrt{2}a)+(2b)+(-3c) \right )\left ((\sqrt{2}a)^{2}+(2b)^{2}+(-3c)^{2}- (\sqrt{2}a)(2b)-(2b)(-3c) -(\sqrt{2}a)(-3c)\right )\\     

= (\sqrt{2}a+2b-3c)(2a^{2}+4b^{2}+9c^{2}-2\sqrt{2}ab+6bc+3\sqrt{2}ca)

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Find the following product (2x-y+3z)(4x^{2}+y^{2}+9z^{2}+2xy+3yz-6xz)

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

Solution

Given:  (2x-y+3z)(4x^{2}+y^{2}+9z^{2}+2xy+3yz-6xz)

This can be written as

(2x-y+3z)((2x)^{2}+(y)^{2}+(3z)^{2}-(2x)(-y)-(-y)(3z)-(2x)(3z))

We know that

a^{3}+b^{3}+c^{3}-3abc= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

Comparing equation (i) with the above identity we get:

a = 2x, b = -y, c = 3z

Hence,

(2x-y+3z)((2x)^{2}+(y)^{2}+(3z)^{2}-(2x)(-y)-(-y)(3z)-(2x)(3z))\\ =(2x)^{3}+(-y)^{3}+(3z)^{3}-3(2x)(-y)(3z)\\ =8x^{3}-y^{3}+27z^{3}+18xyz

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Factorise:

(i) 1+64x^{3}

(ii) a^{3}-2\sqrt{2}b^{3}

 

(i) (1+4x)(1+16x^{2}-4x)

Given 1+64^{3}

(1)^{3}+(4x)^{3}

we know that  a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-ab)

so (1)^{3}+\left (4x \right )^{3}=(1+4x)((1)^{2}+\left (4x \right )^{2}-(1)\left (4x \right ))

(1+4x)(1+16x^{2}-4x)

(ii) (a-\sqrt{2}b)(a^{2}+2b^{2}+\sqrt{2}ab)

Given a^{3}-2\sqrt{2}b^{3}

This can be written as

(a)^{3}-(\sqrt{2}b)^{3}

we know that  a^{3}-b^{3}=(a+b)(a^{2}+b^{2}+ab)

so (a)^{3}-(\sqrt{2}b)^{3}=(a-\sqrt{2}b)(a^{2}+(\sqrt{2}b)^{2}+\sqrt{2}ab)

(a-\sqrt{2}b)(a^{2}+2b^{2}+\sqrt{2}ab)

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Find the following product

(i) \left (\frac{x}{2}+2y \right )\left ( \frac{x^{2}}{4}-xy+4y^{2} \right )

(ii) (x^{2}-1)(x^{4}+x^{2}+1)

(i)      \frac{x^{3}}{8}+8y^{3}

Given: \left (\frac{x}{2}+2y \right )\left ( \frac{x^{2}}{4}-xy+4y^{2} \right )

=\frac{x}{2}\left ( \frac{x^{2}}{4}-xy+4y^{2} \right )+2y \left ( \frac{x^{2}}{4}-xy+4y^{2} \right )\\ =\frac{x^{3}}{8}-\frac{x^{2}y}{2}+\frac{4xy^{2}}{2}+\frac{2yx^{2}}{4}-2xy^{2}+8y^{3}\\ =\frac{x^{3}}{8}-\frac{x^{2}y}{2}+\frac{x^{2}y}{2}+2xy^{2}-2xy^{2}+8y^{3}\\ =\frac{x^{3}}{8}+8y^{3}

(ii) x^{6}-1

Given (x^{2}-1)(x^{4}+x^{2}+1)

=x^{2}(x^{4}+x^{2}+1)-1(x^{4}+x^{2}+1)\\ =x^{6}+x^{4}+x^{2}-x^{2}-x^{4}-x^{2}-1\\ =x^{6}-1

 

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Factorize the following :

\\(i)1-64a^{3}-12a+48a^{2}\\ (ii)8p^{3}+\frac{12}{5}p^{2}+\frac{6}{25}p+\frac{1}{125}

 

(i)(1−4a)(1−4a)(1−4a)

Given, 1−64a3−12a+48a2

The given equation can be written as:

=(1)3−(4a)3−3(1)2(4a)+3(1)(4a)2           ..........(i)

Now we know that:

  a3−b3−3a2b+3ab2=(a−b)3

Comparing equation (i) with the above identity, we get:

∴(1)3−(4a)3−3(1)2(4a)+3(1)(4a)2

=(1−4a)3

Hence the factorized form is (1−4a)(1−4a)(1−4a)

 

(ii)   \left ( 2p+\frac{1}{5} \right )\left ( 2p+\frac{1}{5} \right )\left ( 2p+\frac{1}{5} \right )

Given, 8p^{3}+\frac{12}{5}p^{2}+\frac{6}{25}p+\frac{1}{125}

The given equation can be written as:

(2p)^{3}+\left ( 3 \times (2p)^{2} \times \frac{1}{5}\right )+\left ( 3 (2p) \left (\frac{1}{5} \right )^{2}\right )+\left ( \frac{1}{5} \right )^{3}\\ =(2p)^{3}+\left ( \frac{1}{5} \right )^{3}+\left ( 3 \times (2p)^{2} \times \frac{1}{5}\right )+\left ( 3 (2p) \left (\frac{1}{5} \right )^{2}\right )\\                                 …(i)

Now we know that:

a^{3}+b^{3}+3a^{2}b+3ab^{2}=(a+b)^{3}

Comparing equation (i) with the above identity, we get:

(2p)^{3}+\left ( \frac{1}{5} \right )^{3}+\left ( 3 \times (2p)^{2} \times \frac{1}{5}\right )+\left ( 3 (2p) \left (\frac{1}{5} \right )^{2}\right )=\left ( 2p+\frac{1}{5} \right )^{3}

Hence the factorized form is\left ( 2p+\frac{1}{5} \right )\left ( 2p+\frac{1}{5} \right )\left ( 2p+\frac{1}{5} \right )         

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Expand the following :

(i)(3a-2b)^{3}

(ii)\left ( \frac{1}{x}+\frac{y}{3} \right )^{3}

(iii)\left ( 4-\frac{1}{3x} \right )^{3}

 

(i)27a^{3}-8b^{2}-54a^{2}b+36ab^{2}

Solution: Given (3a-2b)^{3}

We know that

(a-b)^{3}=a^{3}-b^{3}-3a^{2}b+3ab^{2}

So we get:

(3a-2b)^{3}=(3a)^{3}-(2b)^{3}-3(3a)^{2}(2b)+3(3a)(2b)^{2}

(3a-2b)^{3}=27a^{3}-8b^{2}-54a^{2}b+36ab^{2}

Hence the expanded form is 27a^{3}-8b^{2}-54a^{2}b+36ab^{2}

 

(ii)\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x^{2}}+\frac{y^{2}}{3x}

Solution: Given \left ( \frac{1}{x}+\frac{y}{3} \right )^{3}

We know that

(a+b)^{3}=a^{3}+b^{3}+3a^{2}b+3ab^{2}

So we get:

\left ( \frac{1}{x}+\frac{y}{3} \right )^{3}=\left (\frac{1}{x} \right )^{3}+\left (\frac{y}{3} \right )^{3}+3 \left (\frac{1}{x} \right )^{2}\left (\frac{y}{3} \right )+3\left ( \frac{1}{x} \right )\left (\frac{y}{3} \right )^{2}

=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{3y}{3x^{2}}+\frac{3y^{2}}{9x}

=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x^{2}}+\frac{y^{2}}{3x}

Hence the expanded form is =\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x^{2}}+\frac{y^{2}}{3x}

 

(iii)64-\frac{1}{27x^{3}}-\frac{16}{x}+\frac{4}{3x^{2}}

Solution: Given \left ( 4-\frac{1}{3x} \right )^{3}

We know that

(a-b)^{3}=a^{3}-b^{3}-3a^{2}b+3ab^{2}

So we get:

\left ( 4-\frac{1}{3x} \right )^{3}=(4)^{3}-\left ( \frac{1}{3x} \right )^{3}-3(4)^{2}\left ( \frac{1}{3x} \right )+3(4)\left ( \frac{1}{3x} \right )^{2}
=64-\frac{1}{27x^{3}}-\frac{48}{3x}+\frac{12}{9x^{2}}

=64-\frac{1}{27x^{3}}-\frac{16}{x}+\frac{4}{3x^{2}}

Hence the expanded form is  64-\frac{1}{27x^{3}}-\frac{16}{x}+\frac{4}{3x^{2}}.

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