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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)

Or

Length = (2a-1)

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)

Or

Length = (2a-1)

#### 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}

(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

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

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

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

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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

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

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

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

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

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

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

#### 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  $64-\frac{1}{27x^{3}}-\frac{16}{x}+\frac{4}{3x^{2}}$.