NCERT Solutions Class 8 Maths Chapter 8 Algebraic Expressions and Identities Exercise 8.3

In NCERT Solutions Class 8 Maths Chapter 8 Algebraic Expressions and Identities Exercise 8.3, students will learn how to apply special identities to simplify and solve algebraic expressions easily. These identities act like shortcuts that make calculations faster and simpler. This part of the chapter focuses on using these identities to expand and factor expressions.

Our NCERT Solutions for Exercise 8.3 are designed to make each question easier to understand and solve. These solutions are prepared according to the latest NCERT textbook guidelines and help students build a strong foundation. Whether you are revising for exams or just practising, these step-by-step solutions will guide you throughout

1.0Download NCERT Solutions Class 8 Maths Chapter 8 Algebraic Expressions and Identities Exercise 8.3: Free PDF

Download the free NCERT Solutions Class 8 Maths Chapter 8 Algebraic Expressions and Identities Exercise 8.3PDF with step-by-step answers to boost your understanding and exam prep.

2.0Key Concepts in Exercise 8.3 of Class 8 Maths Chapter 8

Exercise 8.3 helps students understand the multiplication of algebraic expressions, an important concept that builds the base for factorization and algebraic identities covered in later parts of the chapter.

Multiplying a Monomial by a Monomial
Students learn to multiply two single-term algebraic expressions (monomials) by multiplying their numerical coefficients and adding the exponents of like variables.

Laws of Exponents for Multiplication
The exercise makes use of the rule:
$x^m\times x^n=x^{m+n}$, where the base is the same.

Multiplying Monomials with Different Variables
Even if variables are different, students learn to multiply all components.

Multiplying More Than Two Monomials
The concept is extended to three or more terms.

Simplification and Rearrangement
Students practice rearranging variables and constants to express the final result in standard form.

3.0NCERT Class 8 Maths Chapter 8: Other Exercises

4.0NCERT Class 8 Maths Chapter 8 Exercise 8.3: Detailed Solutions

1. Carry out the multiplication of the expressions in each of the following pairs: (i) $4p,q+r$ (ii) ab, a - b (iii) $a+b,7a^2{b}^2$ (iv) $a^2-9,4a$ (v) $\mathrm{pq}+\mathrm{qr}+\mathrm{rp},0$ Sol. (i) $4p\times (q+r)=4p\times q+4p\times r$ $=4\mathrm{pq}+4\mathrm{pr}$ (ii) $ab\times (a-b)=ab\times a-ab\times (b)$ $=a^{2}b-a{b}^2$ (iii) $(a+b)\times \left(7a^2{b}^2\right)=7a^2{b}^2\times a+7a^2{b}^2\times b$ $=7{\mathrm{a}}^3{\mathrm{b}}^2+7{\mathrm{a}}^2{\mathrm{b}}^3$ (iv) $\left(a^2-9\right)\times 4a=a^2\times 4a-9\times 4a$ $=4a^3-36a$ (v) $(\mathrm{pq}+\mathrm{qr}+\mathrm{rp})\times 0=0$
2. Complete the table:

Sol. (i) $a(b+c+d)=a\times b+a\times c+a\times d=ab$ $+ac+ad$ (ii) $(x+y-5)\times (5xy)$ $=5xy\times x+5xy\times y+5xy\times (-5)$ $=5x^{2}y+5x{y}^2-25xy$ (iii) $6p^3-7p^2+5p$ (iv) $4p^4{q}^2-4p^2{q}^4$ (v) $a^{2}bc+a{b}^{2}c+ab{c}^2$

3. Find the product: (i) $\left({\mathrm{a}}^2\right)\times \left(2{\mathrm{a}}^{22}\right)\times \left(4{\mathrm{a}}^{26}\right)$ (ii) $\left(\frac{2}{3}xy\right)\times \left(\frac{-9}{10}{x}^2{y}^2\right)$ (iii) $\left(-\frac{10}{3}{\mathrm{pq}}^3\right)\times \left(\frac{6}{5}{\mathrm{p}}^3\mathrm{q}\right)$ (iv) $x\times x^2\times x^3\times x^4$ Sol. (i) $\left({\mathrm{a}}^2\right)\times \left(2{\mathrm{a}}^{22}\right)\times \left(4{\mathrm{a}}^{26}\right)=(1\times 2\times 4)\times$ $\left(a^2\times a^{22}\times a^{26}\right)=8a^2+22+26=8a^{50}$ (ii) $\left(\frac{2}{3}xy\right)\times \left(\frac{-9}{10}{x}^2{y}^2\right)$ $=\left(\frac{2}{3}\times \frac{-9}{10}\right)\times \left(x\times x^2\times y\times y^2\right)$ $=-\frac{3}{5}\times x^{1+2}\times y^{1+2}$ $=-\frac{3}{5}{\mathrm{x}}^3{\mathrm{y}}^3$ (iii) $\left(-\frac{10}{3}{\mathrm{pq}}^3\right)\times \left(\frac{6}{5}{\mathrm{p}}^3\mathrm{q}\right)$ $=\left(-\frac{10}{3}\times \frac{6}{5}\right)\times \left(\mathrm{p}\times {\mathrm{q}}^3\times {\mathrm{p}}^3\times \mathrm{q}\right)$ $=(-4)\times \left(p^4\times q^4\right)=-4p^4{q}^4$ (iv) $\mathrm{x}\times {\mathrm{x}}^2\times {\mathrm{x}}^3\times {\mathrm{x}}^4={\mathrm{x}}^{1+2+3+4}={\mathrm{x}}^{10}$
4. (a) Simplify: $3x(4x-5)+3$ and find its values for (i) $\mathrm{x}=3$, (ii) $\mathrm{x}=\frac{1}{2}$. (b) Simplify: $a\left(a^2+a+1\right)+5$ and find its value for (i) $\mathrm{a}=0$, (ii) $\mathrm{a}=1$, (iii) $\mathrm{a}=-1$. Sol. (a) We have, $3\mathrm{x}(4\mathrm{x}-5)+3=3\mathrm{x}\times 4\mathrm{x}-3\mathrm{x}\times 5+3$ $=12x^2-15x+3$ (i) When $x=3$, then $12(3{)}^2-15(3)+3=66$ (ii) When $\mathrm{x}=\frac{1}{2}$, then $$\begin{gathered} 12{\left(\frac{1}{2}\right)}^2-15\times \frac{1}{2}+3 \\ =3-\frac{15}{2}+3=-\frac{3}{2} \end{gathered}$$ (b) We have, $a\left(a^2+a+1\right)+5$ $=a\times a^2+a\times a+a\times 1+5$ $=a^3+a^2+a+5$ (i) When $a=0$, then $a^3+a^2+a+5$ $=0^3+0^2+0+5$ $=5$ (ii) When $a=1$, then $a^3+a^2+a+5$ $1^3+1^2+1+5$ $=1+1+1+5$ $=3+5=8$ (iii) When $\mathrm{a}=-1$, then ${\mathrm{a}}^3+{\mathrm{a}}^2+\mathrm{a}+5$ $=(-1{)}^3+(-1{)}^2+(-1)+5$ $=-1+1-1+5=4$
5. (a) Add: $\mathrm{p}(\mathrm{p}-\mathrm{q}),\mathrm{q}(\mathrm{q}-\mathrm{r})$ and $\mathrm{r}(\mathrm{r}-\mathrm{p})$ (b) Add: $2\mathrm{x}(\mathrm{z}-\mathrm{x}-\mathrm{y})$ and $2\mathrm{y}(\mathrm{z}-\mathrm{y}-\mathrm{x})$ (c) Subtract: $3\ell (\ell -4m+5n)$ from $4\ell (10n-3m+2\ell )$ (d) Subtract: $3\mathrm{a}(\mathrm{a}+\mathrm{b}+\mathrm{c})-2\mathrm{b}(\mathrm{a}-\mathrm{b}+\mathrm{c})$ from $4c(-a+b+c)$ Sol. (a) $p(p-q)+q(q-r)+r(r-p)$ $=p\times p-p\times q+q\times q-q\times r+r\times r$ $-\mathrm{r}\times \mathrm{p}$ $=p^2-pq+q^2-qr+r^2-pr$ $=p^2+q^2+r^2-pq-qr-pr$ (b) $2x(z-x-y)+2y(z-y-x)$ $=(2x\times z)-(2x\times x)-(2x\times y)+(2y\times z)$ $-(2y\times y)-(2y\times x)$ $=2\mathrm{xz}-2{\mathrm{x}}^2-2\mathrm{xy}+2\mathrm{yz}-2{\mathrm{y}}^2-2\mathrm{xy}$ $=-2x^2-2y^2-4xy+2xz+2yz$. (c) $4\ell (10n-3m+2\ell )-3\ell (\ell -4m+5n)$ $=(4\ell \times 10\mathrm{n})-(4\ell \times 3\mathrm{m})+(4\ell \times 2\ell )+$ $(-3\ell \times \ell )+(-3\ell \times -4\mathrm{m})+(-3\ell \times 5\mathrm{n})$ $=40\ell \mathrm{n}-12\ell \mathrm{m}+8{\ell }^2-3{\ell }^2+12\ell \mathrm{m}-15\ell \mathrm{n}$ $=5{\ell }^2+25\ell n$ (d) $4\mathrm{c}(-\mathrm{a}+\mathrm{b}+\mathrm{c})-\{3\mathrm{a}(\mathrm{a}+\mathrm{b}+\mathrm{c})-2\mathrm{b}(\mathrm{a}-\mathrm{b}+\mathrm{c})\}$ $=\{4c\times (-a)+4c\times b+4c\times c\}-[\{(3a$ $\times a+3a\times b+3a\times c)\}+\{(-2b\times a)+$ $(-2b\times -b)+(-2b\times c)\}]$ $=-4ac+4bc+4c^2-[3a^2+3ab+3ac$ $-2ab+2b^2-2bc]$ $=-4ac+4bc+4c^2-3a^2-3ab-3ac+$ $2\mathrm{ab}-2{\mathrm{b}}^2+2\mathrm{bc}$ $=-3a^2-2b^2+4c^2-7ac+6bc-ab$

5.0Key Features and Benefits of Class 8 Maths Chapter 8 Exercise 8.3

• Focus on Algebraic Identities: In Exercise 8.3, students will learn how to use significant algebraic identities in order to easily expand and simplify expressions. 
• Using Identities will save you time: Using the identities will help students learn to solve complex problems much faster. 
• Step by Step learning: The questions build up so students will feel confident using the identities through the step by step learning process.
• Later Use in Classes: The identities are useful tools when students learn more difficult topics in algebra and will be part of their preparation.
• Follow NCERT Guidelines: All problems are directly based on the official NCERT textbook, ensuring accurate and exam-relevant practice.