Get detailed Rbse Solutions for Class 9 Maths Chapter 2 Exercise 2.4 covering all Algebraic Identities: $(x+y+z)^2$, cube expansion $(a\pm b)^3$, $x^3\pm y^3$ factorisation, and finding dimensions from Area/Volume. Essential for exam preparation.
1. Use suitable identities to find the following products:
(i) $(x + 4)(x + 10)$
Identity: $(x+a)(x+b) = x^2 + (a+b)x + ab$
$$x^2 + (4 + 10)x + (4 \times 10)$$
$$= \mathbf{x^2 + 14x + 40}$$
(ii) $(x + 8)(x – 10)$
Identity: $(x+a)(x+b) = x^2 + (a+b)x + ab$ (Here $a=8, b=-10$)
$$x^2 + (8 + (-10))x + (8 \times -10)$$
$$= x^2 – 2x – 80$$
(iii) $(3x + 4)(3x – 5)$
Identity: $(y+a)(y+b) = y^2 + (a+b)y + ab$ (Here $y=3x, a=4, b=-5$)
$$(3x)^2 + (4 + (-5))(3x) + (4 \times -5)$$
$$= 9x^2 + (-1)(3x) – 20$$
$$= \mathbf{9x^2 – 3x – 20}$$
(iv) $(y^2 + \frac{3}{2})(y^2 – \frac{3}{2})$
Identity: $(a+b)(a-b) = a^2 – b^2$ (Here $a=y^2, b=\frac{3}{2}$)
$$(y^2)^2 – \left(\frac{3}{2}\right)^2$$
$$= \mathbf{y^4 – \frac{9}{4}}$$
(v) $(3 – 2x)(3 + 2x)$
Identity: $(a-b)(a+b) = a^2 – b^2$ (Here $a=3, b=2x$)
$$(3)^2 – (2x)^2$$
$$= \mathbf{9 – 4x^2}$$
2. Evaluate the following products without multiplying directly:
(i) $103 \times 107$
$$103 \times 107 = (100 + 3)(100 + 7)$$
Identity: $(x+a)(x+b) = x^2 + (a+b)x + ab$
$$= (100)^2 + (3+7)(100) + (3 \times 7)$$
$$= 10000 + 10(100) + 21$$
$$= 10000 + 1000 + 21 = \mathbf{11021}$$
(ii) $95 \times 96$
$$95 \times 96 = (100 – 5)(100 – 4)$$
Identity: $(x+a)(x+b) = x^2 + (a+b)x + ab$
$$= (100)^2 + (-5 + (-4))(100) + (-5 \times -4)$$
$$= 10000 + (-9)(100) + 20$$
$$= 10000 – 900 + 20 = \mathbf{9120}$$
(iii) $104 \times 96$
$$104 \times 96 = (100 + 4)(100 – 4)$$
Identity: $(a+b)(a-b) = a^2 – b^2$
$$= (100)^2 – (4)^2$$
$$= 10000 – 16 = \mathbf{9984}$$
3. Factorise the following using appropriate identities:
(i) $9x^2 + 6xy + y^2$
Identity: $a^2 + 2ab + b^2 = (a+b)^2$
$$= (3x)^2 + 2(3x)(y) + (y)^2$$
$$= \mathbf{(3x + y)^2}$$
(ii) $4y^2 – 4y + 1$
Identity: $a^2 – 2ab + b^2 = (a-b)^2$
$$= (2y)^2 – 2(2y)(1) + (1)^2$$
$$= \mathbf{(2y – 1)^2}$$
(iii) $x^2 – \frac{y^2}{100}$
Identity: $a^2 – b^2 = (a-b)(a+b)$
$$= (x)^2 – \left(\frac{y}{10}\right)^2$$
$$= \mathbf{\left(x – \frac{y}{10}\right)\left(x + \frac{y}{10}\right)}$$
4. Expand each of the following, using suitable identities:
Identity: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$
(i) $(x + 2y + 4z)^2$
$$(x)^2 + (2y)^2 + (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)$$
$$= \mathbf{x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx}$$
(ii) $(2x – y + z)^2$
$$(2x)^2 + (-y)^2 + (z)^2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)$$
$$= \mathbf{4x^2 + y^2 + z^2 – 4xy – 2yz + 4zx}$$
(iii) $(–2x + 3y + 2z)^2$
$$(-2x)^2 + (3y)^2 + (2z)^2 + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)$$
$$= \mathbf{4x^2 + 9y^2 + 4z^2 – 12xy + 12yz – 8zx}$$
(iv) $(3a – 7b – c)^2$
$$(3a)^2 + (-7b)^2 + (-c)^2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)$$
$$= \mathbf{9a^2 + 49b^2 + c^2 – 42ab + 14bc – 6ca}$$
(v) $(–2x + 5y – 3z)^2$
$$(-2x)^2 + (5y)^2 + (-3z)^2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)$$
$$= \mathbf{4x^2 + 25y^2 + 9z^2 – 20xy – 30yz + 12zx}$$
(vi) 12$\left[\frac{1}{4}a – \frac{1}{2}b + 1\right]^2$
$$\left(\frac{1}{4}a\right)^2 + \left(-\frac{1}{2}b\right)^2 + (1)^2 + 2\left(\frac{1}{4}a\right)\left(-\frac{1}{2}b\right) + 2\left(-\frac{1}{2}b\right)(1) + 2(1)\left(\frac{1}{4}a\right)$$
$$= \frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 – \frac{1}{4}ab – b + \frac{1}{2}a$$
5. Factorise:
(i) $4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz$
This matches the expanded form $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a+b+c)^2$.
Identify $a=2x, b=3y, c=4z$.
Since the $yz$ and $xz$ terms are negative, the variable common to both ($z$) must be negative. Let $c=-4z$.
$$= (2x)^2 + (3y)^2 + (-4z)^2 + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)$$
$$= \mathbf{(2x + 3y – 4z)^2}$$
(ii) $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz$
Since the $x$ terms ($xy$ and $xz$) are negative, we set $a = -\sqrt{2}x$.
$$= (-\sqrt{2}x)^2 + (y)^2 + (2\sqrt{2}z)^2 + 2(-\sqrt{2}x)(y) + 2(y)(2\sqrt{2}z) + 2(2\sqrt{2}z)(-\sqrt{2}x)$$
$$= \mathbf{(-\sqrt{2}x + y + 2\sqrt{2}z)^2}$$
6. Write the following cubes in expanded form:
Identities: $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ and $(a-b)^3 = a^3 – b^3 – 3ab(a-b)$
(i) $(2x + 1)^3$
$$= (2x)^3 + (1)^3 + 3(2x)(1)(2x + 1)$$
$$= 8x^3 + 1 + 6x(2x + 1)$$
$$= \mathbf{8x^3 + 12x^2 + 6x + 1}$$
(ii) $(2a – 3b)^3$
$$= (2a)^3 – (3b)^3 – 3(2a)(3b)(2a – 3b)$$
$$= 8a^3 – 27b^3 – 18ab(2a – 3b)$$
$$= \mathbf{8a^3 – 27b^3 – 36a^2b + 54ab^2}$$
(iii) $\left(\frac{3}{2}x + 1\right)^3$
$$= \left(\frac{3}{2}x\right)^3 + (1)^3 + 3\left(\frac{3}{2}x\right)(1)\left(\frac{3}{2}x + 1\right)$$
$$= \frac{27}{8}x^3 + 1 + \frac{9}{2}x\left(\frac{3}{2}x + 1\right)$$
$$= \mathbf{\frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1}$$
(iv) $\left(x – \frac{2}{3}y\right)^3$
$$= (x)^3 – \left(\frac{2}{3}y\right)^3 – 3(x)\left(\frac{2}{3}y\right)\left(x – \frac{2}{3}y\right)$$
$$= x^3 – \frac{8}{27}y^3 – 2xy\left(x – \frac{2}{3}y\right)$$
$$= \mathbf{x^3 – 2x^2y + \frac{4}{3}xy^2 – \frac{8}{27}y^3}$$
7. Evaluate the following using suitable identities:
(i) $(99)^3$
$$= (100 – 1)^3$$
Identity: $(a-b)^3 = a^3 – 3a^2b + 3ab^2 – b^3$
$$= (100)^3 – 3(100)^2(1) + 3(100)(1)^2 – (1)^3$$
$$= 1000000 – 30000 + 300 – 1$$
$$= 970300 – 1 = \mathbf{970299}$$
(ii) $(102)^3$
$$= (100 + 2)^3$$
Identity: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$$= (100)^3 + 3(100)^2(2) + 3(100)(2)^2 + (2)^3$$
$$= 1000000 + 60000 + 1200 + 8$$
$$= \mathbf{1061208}$$
(iii) $(998)^3$
$$= (1000 – 2)^3$$
Identity: $(a-b)^3 = a^3 – 3a^2b + 3ab^2 – b^3$
$$= (1000)^3 – 3(1000)^2(2) + 3(1000)(2)^2 – (2)^3$$
$$= 1000000000 – 6000000 + 12000 – 8$$
$$= 994012000 – 8 = \mathbf{994011992}$$
8. Factorise each of the following:
These are factorisations of the expanded cube identities, $(a \pm b)^3$.
(i) $8a^3 + b^3 + 12a^2b + 6ab^2$
$$= (2a)^3 + (b)^3 + 3(2a)^2(b) + 3(2a)(b)^2$$
Identity: $a^3 + b^3 + 3a^2b + 3ab^2 = (a+b)^3$
$$= \mathbf{(2a + b)^3}$$
(ii) $8a^3 – b^3 – 12a^2b + 6ab^2$
$$= (2a)^3 – (b)^3 – 3(2a)^2(b) + 3(2a)(b)^2$$
Identity: $a^3 – b^3 – 3a^2b + 3ab^2 = (a-b)^3$
$$= \mathbf{(2a – b)^3}$$
(iii) $27 – 125a^3 – 135a + 225a^2$
Rearrange into $a^3 – 3a^2b + 3ab^2 – b^3$:
$$= (3)^3 + 3(3)(5a)^2 – 3(3)^2(5a) – (5a)^3$$
$$= (3)^3 – 3(9)(5a) + 3(3)(25a^2) – (5a)^3$$
This is the expansion of $(3 – 5a)^3$.
$$= \mathbf{(3 – 5a)^3}$$
(iv) $64a^3 – 27b^3 – 144a^2b + 108ab^2$
$$= (4a)^3 – (3b)^3 – 3(4a)^2(3b) + 3(4a)(3b)^2$$
This is the expansion of $(4a – 3b)^3$.
$$= \mathbf{(4a – 3b)^3}$$
(v) $27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p$
$$= (3p)^3 – \left(\frac{1}{6}\right)^3 – 3(3p)^2\left(\frac{1}{6}\right) + 3(3p)\left(\frac{1}{6}\right)^2$$
$$= (3p)^3 – \left(\frac{1}{6}\right)^3 – 3(9p^2)\left(\frac{1}{6}\right) + 9p\left(\frac{1}{36}\right)$$
$$= (3p)^3 – \left(\frac{1}{6}\right)^3 – \frac{9}{2}p^2 + \frac{1}{4}p$$
This matches the expansion of $\left(3p – \frac{1}{6}\right)^3$.
$$= \mathbf{\left(3p – \frac{1}{6}\right)^3}$$
9. Verify:
(i) $x^3 + y^3 = (x + y)(x^2 – xy + y^2)$
RHS (Right Hand Side):
$$(x + y)(x^2 – xy + y^2)$$
$$= x(x^2 – xy + y^2) + y(x^2 – xy + y^2)$$
$$= x^3 – x^2y + xy^2 + x^2y – xy^2 + y^3$$
The middle four terms cancel out:
$$= x^3 + y^3$$
Since RHS = LHS, the identity is verified.
(ii) $x^3 – y^3 = (x – y)(x^2 + xy + y^2)$
RHS:
$$(x – y)(x^2 + xy + y^2)$$
$$= x(x^2 + xy + y^2) – y(x^2 + xy + y^2)$$
$$= x^3 + x^2y + xy^2 – x^2y – xy^2 – y^3$$
The middle four terms cancel out:
$$= x^3 – y^3$$
Since RHS = LHS, the identity is verified.
10. Factorise each of the following:
These use the identities verified in Q9.
(i) $27y^3 + 125z^3$
$$= (3y)^3 + (5z)^3$$
Identity: $a^3 + b^3 = (a + b)(a^2 – ab + b^2)$
$$= \mathbf{(3y + 5z)((3y)^2 – (3y)(5z) + (5z)^2)}$$
$$= \mathbf{(3y + 5z)(9y^2 – 15yz + 25z^2)}$$
(ii) $64m^3 – 343n^3$
$$= (4m)^3 – (7n)^3$$
Identity: $a^3 – b^3 = (a – b)(a^2 + ab + b^2)$
$$= \mathbf{(4m – 7n)((4m)^2 + (4m)(7n) + (7n)^2)}$$
$$= \mathbf{(4m – 7n)(16m^2 + 28mn + 49n^2)}$$
11. Factorise: $27x^3 + y^3 + z^3 – 9xyz$
Identity: $a^3 + b^3 + c^3 – 3abc = (a+b+c)(a^2 + b^2 + c^2 – ab – bc – ca)$
Here, $a=3x, b=y, c=z$.
$$= (3x)^3 + (y)^3 + (z)^3 – 3(3x)(y)(z)$$
$$= (3x + y + z)((3x)^2 + y^2 + z^2 – (3x)y – yz – z(3x))$$
$$= \mathbf{(3x + y + z)(9x^2 + y^2 + z^2 – 3xy – yz – 3zx)}$$
12. Verify that $x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x+y+z)[(x-y)^2 + (y-z)^2 + (z-x)^2]$
We start with the RHS and simplify the square terms:
$$(x-y)^2 + (y-z)^2 + (z-x)^2$$
$$= (x^2 – 2xy + y^2) + (y^2 – 2yz + z^2) + (z^2 – 2zx + x^2)$$
$$= 2x^2 + 2y^2 + 2z^2 – 2xy – 2yz – 2zx$$
Factor out 2:
$$= 2(x^2 + y^2 + z^2 – xy – yz – zx)$$
Substitute this back into the RHS:
$$\text{RHS} = \frac{1}{2}(x+y+z)[2(x^2 + y^2 + z^2 – xy – yz – zx)]$$
The $\frac{1}{2}$ and $2$ cancel out:
$$\text{RHS} = (x+y+z)(x^2 + y^2 + z^2 – xy – yz – zx)$$
This is the standard expansion form for the LHS identity:
$$\text{RHS} = x^3 + y^3 + z^3 – 3xyz$$
Since RHS = LHS, the identity is verified.
13. If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.
Use the identity from Question 12 (or 11):
$$x^3 + y^3 + z^3 – 3xyz = (x+y+z)(x^2 + y^2 + z^2 – xy – yz – zx)$$
We are given the condition that $x + y + z = 0$. Substitute this into the RHS:
$$x^3 + y^3 + z^3 – 3xyz = (0)(x^2 + y^2 + z^2 – xy – yz – zx)$$
$$x^3 + y^3 + z^3 – 3xyz = 0$$
Transpose the $3xyz$ term:
$$\mathbf{x^3 + y^3 + z^3 = 3xyz}$$
The statement is shown.
14. Without actually calculating the cubes, find the value of each of the following:
Use the result from Q13: If $a+b+c = 0$, then $a^3 + b^3 + c^3 = 3abc$.
(i) $(-12)^3 + (7)^3 + (5)^3$
Check the sum of the bases: $a+b+c = -12 + 7 + 5 = 0$.
Since the sum is zero:
$$(-12)^3 + (7)^3 + (5)^3 = 3(-12)(7)(5)$$
$$= 3(-420) = \mathbf{-1260}$$
(ii) $(28)^3 + (-15)^3 + (-13)^3$
Check the sum of the bases: $a+b+c = 28 + (-15) + (-13) = 28 – 28 = 0$.
Since the sum is zero:
$$(28)^3 + (-15)^3 + (-13)^3 = 3(28)(-15)(-13)$$
$$= 3(5460) = \mathbf{16380}$$
15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
We factorise the quadratic area expression using splitting the middle term.
(i) Area: $25a^2 – 35a + 12$
Product $a \cdot c = 25 \times 12 = 300$. Sum $b = -35$.
Numbers: $-15$ and $-20$.
$$25a^2 – 15a – 20a + 12$$
$$5a(5a – 3) – 4(5a – 3)$$
$$= (5a – 3)(5a – 4)$$
Possible expressions for Length and Breadth are: $\mathbf{(5a – 3)}$ and $\mathbf{(5a – 4)}$.
(ii) Area: $35y^2 + 13y – 12$
Product $a \cdot c = 35 \times -12 = -420$. Sum $b = 13$.
Numbers: $28$ and $-15$.
$$35y^2 + 28y – 15y – 12$$
$$7y(5y + 4) – 3(5y + 4)$$
$$= (5y + 4)(7y – 3)$$
Possible expressions for Length and Breadth are: $\mathbf{(5y + 4)}$ and $\mathbf{(7y – 3)}$.
16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
Volume of a cuboid = Length $\times$ Breadth $\times$ Height. We need three factors.
(i) Volume: $3x^2 – 12x$
Factor out the common term, $3x$:
$$3x(x – 4)$$
Possible expressions for the dimensions are: $\mathbf{3}$, $\mathbf{x}$, and $\mathbf{(x – 4)}$.
(ii) Volume: $12ky^2 + 8ky – 20k$
First, factor out the common term, $4k$:
$$4k(3y^2 + 2y – 5)$$
Now, factor the quadratic $3y^2 + 2y – 5$ (Product $-15$, Sum $2$. Numbers: $5$ and $-3$):
$$3y^2 + 5y – 3y – 5$$
$$y(3y + 5) – 1(3y + 5)$$
$$= (3y + 5)(y – 1)$$
Substitute back into the volume expression:
$$4k(3y + 5)(y – 1)$$
Possible expressions for the dimensions are: $\mathbf{4k}$, $\mathbf{(3y + 5)}$, and $\mathbf{(y – 1)}$.