Latest Updated : November 2023

If you are in search of **NCERT solutions for Class 8 Maths Chapter** **8 Algebraic Expressions and Identities**, then you are at the right place. When it comes to **Mathematics**, it is a subject that holds value for students across various academic streams, be it science, biology, or commerce and having a very good understanding of basic math concepts will help you in achieving your goal and make it much more easier.

Here we provide detailed **NCERT solutions for Class 8 Maths**Â of all the chapters,Â **exercise wise**Â withÂ **updated syllabus by NCERT**. These NCERT solutions for Class 8 Maths have been carefully crafted to provide detailed explanations, ensuring that students can grasp the concepts effectively. We also provide extra practice questions for the same, so feel free to bookmarkÂ **ncertforclass8.com**Â to boost your preparation.

## NCERT Solutions for Class 8 Maths Chapter 8 **Algebraic Expressions and Identities** Ex 8.3

**Class 8 Maths Chapter** **8 Algebraic Expressions and Identities** **Exercise **8.3

**Exercise 8.3**

**1. Carry out the multiplication of the expressions in each of the following pairs.**

`\[(i)\ 4p\ ,\ q+r\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ii)\ ab,\ a-b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (iii)\ a+b,\ 7a^2b^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (iv)\ a^2-9\ ,\ 4a\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (v)\ pq+qr+rp,\ 0\]`

Solution :

`\[(i)\ 4p(q+r)=4pq+4pr\]`

`\[(ii)\ ab(a-b)=a^2b-ab^2\]`

`\[(iii)\ (a+b)\ (7a^2b^2\ )=7a^3b^2+7a^2b^3\]`

`\[(iv)\ (a^2-9)(4a)=4a^3-36a\]`

`\[(v)\ (pq+qr+rp)\times0=0\]`

**2. Complete the table.**

Solution:

**3. Find the product.**

`\[\left(i\right)\ \left(a^2\right)\ \times\ \left(2a^{22}\right)\ \times\ \left(4a^{26}\right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(ii\right)\ \left(\frac{2}{3}xy\right)\ \times\ \left(-\frac{9}{10}x^2y^2\right)\]`

`\[\left(iii\right)\ \left(-\frac{10}{3}pq^3\right)\ \times\ \left(\frac{6}{5}p^3q\right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(iv\right)\ x\ \times\ x^2\ \times\ x^3\ \times\ x^4\]`

Solution :

`\[\left(i\right)\ \left(a^2\right)\ \times\ \left(2a^{22}\right)\ \times\ \left(4a^{26}\right)\]`

`\[=(2\times4)\ \times(a^2\times a^{22}\times a^{26})\]`

`\[=8\times\left(a^{2+22+26}\right)\]`

`\[=8a^{50}\]`

`\[\ \left(ii\right)\ \left(\frac{2}{3}xy\right)\ \times\ \left(-\frac{9}{10}x^2y^2\right)\]`

`\[=\left(\ \frac{2}{3}\times-\frac{9}{10})\right)\times\ (x\times x^2\times y\times y^2)\]`

`\[=(-\frac{3}{5}x^3y^3)\]`

`\[\left(iii\right)\ \left(-\frac{10}{3}pq^3\right)\ \times\ \left(\frac{6}{5}p^3q\right)\]`

`\[=(-\frac{10}{3}\times\frac{6}{5})\times(p\times p^3\times q^3\times q)\]`

`\[=(-4p^4q^4)\]`

`\[\left(iv\right)\ x\ \times\ x^2\ \times\ x^3\ \times\ x^4\]`

`\[=x^{1+2+3+4}\]`

`\[=x^{10}\]`

**4. (a) Simplify 3x (4x â€“ 5) + 3 and find its values for **

**(i) x = 3 (ii) x =1/2**

Solution :

`\[a)\ 3x\ (4x-5)+3\]`

`\[=3x\ (4x)-3x\ (5)+3\]`

`\[=12x^2-15x+3\]`

`\[(i)Putting\ x=3\ \ in\ the\ equation\ \ 12x^2-15x+3\ we\ get\]`

`\[=12(32)-15(3)+3\]`

`\[=108\ -\ 45\ +\ 3\ =\ 66\]`

`\[(ii)Putting\ x=\frac{1}{2}\ \ in\ the\ equation\ 12x^2-15x+3\ we\ get\]`

`\[=12\left(\frac{1}{2}\right)^2-15\left(\frac{1}{2}\right)+3\]`

`\[=12\left(\frac{1}{4}\right)-\frac{15}{2}+3\]`

`\[=3-\frac{15}{2}+3\]`

`\[=6-\frac{15}{2}=\ \frac{\left(12-15\right)}{2}\]`

`\[=-3/2\]`

`\[\left(b\right)\ Simplify\ a\left(a^2\ +\ a\ +\ 1\right)\ +5\ and\ find\ its\ values\ for\ \ \left(i\right)\ a=0\ \ ,\ \ \ \left(ii\right)\ a\ =1,\ \ \left(iii\right)\ a\ =\ -1\]`

Solution:

`\[=a\times a^2+a\times a+a\times1+5\]`

`\[=a^3+a^2+a+5\]`

`\[(i)\ putting\ a=0\ in\ the\ equation\ we\ get\ =\ 0^3+0^2+0+5=\ 5\]`

`\[(ii)\ putting\ a=1\ in\ the\ equation\ we\ get\ =1^3+1^2+1+5=1+1+1+5=8\]`

`\[(iii)\ Putting\ a=-1\ in\ the\ equation\ we\ get\ =(-1)^3+(-1)^2+(-1)+5=-1+1-1+5=4\]`

**5. (a) Add: p ( p â€“ q), q ( q â€“ r) and r ( r â€“ p)**

Solution :

`\[a)\ p(p-q)+q(q-r)+r(r-p)\]`

`\[=(p^2-pq)+(q^2-qr)+(r^2-pr)\]`

`\[=p^2+q^2+r^2-pq-qr-pr\]`

**(b) Add: 2x (z â€“ x â€“ y) and 2y (z â€“ y â€“ x)**

Solution :

`\[b)\ 2x\ (z-x-y)+2y\ (z-y-x)\]`

`\[=(2xz-2x^2-2xy)+(2yz-2y^2-2xy)\]`

`\[=2xz-4xy+2yz-2x^2-2y^2\]`

**(c) Subtract: 3l (l â€“ 4 m + 5 n) from 4l ( 10 n â€“ 3 m + 2 l )**

Solution :

`\[c)\ 4l\ (10n-3m+2l)-3l(l-4m+5n)\]`

`\[=(40\ln-12lm+8l^2)-(3l^2-12lm+15\ln)\]`

`\[=40\ln-12lm+8l^2-3l^2+12lm-15\ln\]`

`\[=25\ln+5l^2\]`

**(d) Subtract: 3a (a + b + c ) â€“ 2 b (a â€“ b + c) from 4c ( â€“ a + b + c )**

Solution :

`\[d)\ 4c\ (-a+b+c)-3a\ (a+b+c)-2b(a-b+c)\]`

`\[=(-4ac+4bc+4c^2)-(3a^2+3ab+3ac-(2ab-2b^2+2bc))\]`

`\[=-4ac+4bc+4c^2-(3a^2+3ab+3ac-2ab+2b^2-2bc)\]`

`\[=-4ac+4bc+4c^2-3a^2-3ab-3ac+2ab-2b^2+2bc\]`

`\[=-7ac+6bc+4c^2-3a^2-ab-2b^2\]`

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