Q&A - Ask Doubts and Get Answers

#2.3

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)

View Full Answer(1)

Posted by

infoexpert24

#2.3

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.

View Full Answer(1)

Posted by

infoexpert24

#2.3

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

View Full Answer(1)

Posted by

infoexpert24

#2.3

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.

View Full Answer(1)

Posted by

infoexpert24

#2.3

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

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

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

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

View Full Answer(1)

Posted by

infoexpert24

#2.3

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,

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

View Full Answer(1)

Posted by

infoexpert24

#2.3

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

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

View Full Answer(1)

Posted by

infoexpert24

#2.3

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

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

View Full Answer(1)

Posted by

infoexpert24

#2.3

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

View Full Answer(1)

Posted by

infoexpert24

#2.3

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

View Full Answer(1)

Exams

Colleges

Predictors

Resources

Quick links

Quick Links

Exams

Colleges

Resources

Colleges

Resources

Exams

Exams

Colleges

Resources

Diploma Colleges

Other Exams

Exams

Resources

Upcoming Events

Exams

Ranking

Products & Resources

NCERT Solutions

Top Countries

Student Visas

Exams

Colleges

Exams

Colleges

Upcoming Events/Predictors

Resources

Exams

Colleges & Courses

Predictors

Resources

Online Courses

Products

Engineering Preparation

Medical Preparation

Top Streams

Specializations

Resources

Top Providers

Exams

Colleges

Predictors

Resources

Resources

Exams

Colleges

Predictors & E-Books

Exams

Animation Courses

Colleges

Resources

Exams

Colleges

Resources

Top Courses & Careers

Colleges

Exams

Resources

Get Answers to all your Questions

All Questions

Give possible expressions for the length and breadth of the rectangle whose area is given by

Posted by

infoexpert24

Find the value of (i) when (ii) when

Posted by

infoexpert24

Crack CUET with india's "Best Teachers"

Without finding the cubes, factorize

Posted by

infoexpert24

Without actually calculating the cubes, find the value of (i) (ii)

Posted by

infoexpert24

Crack NEET with "AI Coach"

Factorize (i) (ii)

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

Find the value of

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

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

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

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

Factorize

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

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

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

Factorise:

(i) $1+64x^{3}$

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

Find the following product

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

(ii) $(x^{2}-1)(x^{4}+x^{2}+1)$

Factorize the following :

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

Expand the following :

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

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

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