NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4 - Polynomials

Q1 (i) Use suitable identities to find the following product: $(x + 4) ( x + 10)$

Answer:

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put $a = 4 \ \ and \ \ b = 10$

$(x+4)(x+10)= x^2+(4 + 10)x+4\times 10$

$= x^2+14x+40$

Therefore, $(x + 4) ( x + 10)$ is equal to $x^2+14x+40$

Q1 (ii) Use suitable identities to find the following product: $(x+8)(x-10)$

Answer:

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put $a = 8 \ \ and \ \ b = -10$

$(x+8)(x-10)= x^2+(8 + (-10))x+(-10)\times 8$

$= x^2-2x-80$

Therefore, $(x+8)(x-10)$ is equal to $x^2-2x-80$

Q1 (iii) Use suitable identities to find the following product: $(3x+4)(3x - 5)$

Answer:

We can write $(3x+4)(3x - 5)$ as

$(3x+4)(3x - 5)= 9\left ( x+\frac{4}{3} \right )\left ( x-\frac{5}{3} \right )$

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put $a = \frac{4}{3} \ \ and \ \ b = -\frac{5}{3}$

$9\left ( x+\frac{4}{3} \right )\left ( x-\frac{5}{3} \right )= 9\left ( x^2+\left ( \frac{4}{3}-\frac{5}{3} \right )x+\frac{4}{3} \times \left ( -\frac{5}{3} \right ) \right )$

$=9x^2-3x-20$

Therefore, $(3x+4)(3x - 5)$ is equal to $9x^2-3x-20$

Q1 (iv) Use suitable identities to find the following product: $(y^2 + \frac{3}{2})(y^2 - \frac{3}{2})$

Answer:

We will use identity

$(x+a)(x-a)=x^2-a^2$

Put $x=y^2 \ \ and \ \ a = \frac{3}{2}$

$(y^2 + \frac{3}{2})(y^2 - \frac{3}{2}) = \left ( y^2 \right )^2-\left(\frac{3}{2} \right )^2$

$= y^4-\frac{9}{4}$

Therefore, $(y^2 + \frac{3}{2})(y^2 - \frac{3}{2})$ is equal to $y^4-\frac{9}{4}$

Q1 (v) Use suitable identities to find the following product: $(3 - 2x) (3 + 2x)$

Answer:

We can write $(3 - 2x) (3 + 2x)$ as

$(3 - 2x) (3 + 2x)=-4\left ( x-\frac{3}{2} \right )\left(x+\frac{3}{2} \right )$

We will use identity

$(x+a)(x-a)=x^2-a^2$

Put $a = \frac{3}{2}$

$-4(x + \frac{3}{2})(x- \frac{3}{2}) =-4\left ( \left ( x \right )^2-\left(\frac{3}{2} \right )^2 \right )$

$=9-4x^2$

Therefore, $(3 - 2x) (3 + 2x)$ is equal to $9-4x^2$

Q2 (i) Evaluate the following product without multiplying directly: $103 \times 107$

Answer:

We can rewrite $103 \times 107$ as

$\Rightarrow 103 \times 107= (100+3)\times (100+7)$

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put $x =100 , a=3 \ \ and \ \ b = 7$

$(100+3)\times (100+7)= (100)^2+(3+7)100+3\times 7$

$=10000+1000+21= 11021$

Therefore, value of $103 \times 107$ is $11021$

Q2 (ii) Evaluate the following product without multiplying directly: $95 \times 96$

Answer:

We can rewrite $95 \times 96$ as

$\Rightarrow 95 \times 96= (100-5)\times (100-4)$

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put $x =100 , a=-5 \ \ and \ \ b = -4$

$(100-5)\times (100-4)= (100)^2+(-5-4)100+(-5)\times (-4)$

$=10000-900+20= 9120$

Therefore, value of $95 \times 96$ is $9120$

Q2 (iii) Evaluate the following product without multiplying directly: $104 \times 96$

Answer:

We can rewrite $104 \times 96$ as

$\Rightarrow 104 \times 96= (100+4)\times (100-4)$

We will use identity

$(x+a)(x-a)=x^2-a^2$

Put $x =100 \ \ and \ \ a=4$

$(100+4)\times (100-4)= (100)^2-(4)^2$

$=10000-16= 9984$

Therefore, value of $104 \times 96$ is $9984$

Q3 (i) Factorise the following using appropriate identities: $9x^2 + 6xy + y^2$

Answer:

We can rewrite $9x^2 + 6xy + y^2$ as

$\Rightarrow 9x^2 + 6xy + y^2 = (3x)^2+2\times 3x\times y +(y)^2$

Using identity $\Rightarrow (a+b)^2 = (a)^2+2\times a\times b +(b)^2$

Here, $a= 3x \ \ and \ \ b = y$

Therefore,

$9x^2+6xy+y^2 = (3x+y)^2 = (3x+y)(3x+y)$

Q3 (ii) Factorise the following using appropriate identities: $4y^2 - 4y + 1$

Answer:

We can rewrite $4y^2 - 4y + 1$ as

$\Rightarrow 4y^2 - 4y + 1=(2y)^2-2\times2y\times 1+(1)^2$

Using identity $\Rightarrow (a-b)^2 = (a)^2-2\times a\times b +(b)^2$

Here, $a= 2y \ \ and \ \ b = 1$

Therefore,

$4y^2 - 4y + 1=(2y-1)^2=(2y-1)(2y-1)$

Q3 (iii) Factorise the following using appropriate identities: $x^2 - \frac{y^2}{100}$

Answer:

We can rewrite $x^2 - \frac{y^2}{100}$ as

$\Rightarrow x^2 - \frac{y^2}{100} = (x)^2-\left(\frac{y}{10} \right )^2$

Using identity $\Rightarrow a^2-b^2 = (a-b)(a+b)$

Here, $a= x \ \ and \ \ b = \frac{y}{10}$

Therefore,

$x^2 - \frac{y^2}{100} = \left ( x-\frac{y}{10} \right )\left ( x+\frac{y}{10} \right )$

Q4 (i) Expand each of the following, using suitable identities: $(x + 2y+4z)^2$

Answer:

Given is $(x + 2y+4z)^2$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a = x , b = 2y \ \ and \ \ c = 4z$

Therefore,

$(x + 2y+4z)^2 = (x)^2+(2y)^2+(4z)^2+2.x.2y+2.2y.4z+2.4z.x$

$= x^2+4y^2+16z^2+4xy+16yz+8zx$

Q4 (ii) Expand each of the following, using suitable identities: $(2x - y + z)^2$

Answer:

Given is $(2x - y + z)^2$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a = 2x , b = -y \ \ and \ \ c = z$

Therefore,

$(2x -y+z)^2 = (2x)^2+(-y)^2+(z)^2+2.2x.(-y)+2.(-y).z+2.z.2x$

$= 4x^2+y^2+z^2-4xy-2yz+4zx$

Q4 (iii) Expand each of the following, using suitable identities: $(-2x + 3y + 2z)^2$

Answer:

Given is $(-2x + 3y + 2z)^2$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a =- 2x , b = 3y \ \ and \ \ c = 2z$

Therefore,

$(-2x +3y+2z)^2 = (-2x)^2+(3y)^2+(2z)^2+2.(-2x).3y+2.3y.2z+2.z.(-2x)$

$= 4x^2+9y^2+4z^2-12xy+12yz-8zx$

Q4 (iv) Expand each of the following, using suitable identities: $(3a - 7b - c)^2$

Answer:

Given is $(3a - 7b - c)^2$

We will Use identity

$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$

Here, $x =3a , y = -7b \ \ and \ \ z = -c$

Therefore,

$(3a - 7b - c)^2=(3a)^2+(-7b)^2+(-c)^2+2.3a.(-7b)+2.(-7b).(-c)+2.(-c)$ $.3a$

$= 9a^2+49b^2+c^2-42ab+14bc-6ca$

Q4 (v) Expand each of the following, using suitable identities: $(-2x + 5y -3z)^2$

Answer:

Given is $(-2x + 5y -3z)^2$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a =- 2x , b = 5y \ \ and \ \ c = -3z$

Therefore,

$(-2x +5y-3z)^2$ $= (-2x)^2+(5y)^2+(-3z)^2+2.(-2x).5y+2.5y.(-3z)+2.(-3z).(-2x)$

$= 4x^2+25y^2+9z^2-20xy-30yz+12zx$

Q4 (vi) Expand each of the following, using suitable identities: $\left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2$

Answer:

Given is $\left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2$

We will Use identity

$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$

Here, $x =\frac{a}{4} , y = -\frac{b}{2} \ \ and \ \ z = 1$

Therefore,

$\left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2$ $=\left(\frac{a}{4} \right )^2+\left(-\frac{b}{2} \right )^2+(1)^2+2.\left(\frac{a}{4} \right ). \left(-\frac{b}{2} \right )+2. \left(-\frac{b}{2} \right ).1+2.1.\left(\frac{a}{4} \right )$

$= \frac{a^2}{16}+\frac{b^2}{4}+1-\frac{ab}{4}-b+\frac{a}{2}$

Q5 (i) Factorise: $4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz$

Answer:

We can rewrite $4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz$ as

$\Rightarrow 4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz$ $= (2x)^2+(3y)^2+(-4z)^2+2.2x.3y+2.3y.(-4z)+2.(-4z).2x$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a = 2x , b = 3y \ \ and \ \ c = -4z$

Therefore,

$4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz = (2x+3y-4z)^2$

$= (2x+3y-4z)(2x+3y-4z)$

Q5 (ii) Factorise: $2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz$

Answer:

We can rewrite $2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz$ as

$\Rightarrow 2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz$ $= (-\sqrt2x)^2+(y)^2+(2\sqrt2z)^2+2.(-\sqrt2).y+2.y.2\sqrt2z+2.(-\sqrt2x).2\sqrt2z$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a = -\sqrt2x , b = y \ \ and \ \ c = 2\sqrt2z$

Therefore,

$2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz=(-\sqrt2x+y+2\sqrt2z)^2$

$=(-\sqrt2x+y+2\sqrt2z)(-\sqrt2x+y+2\sqrt2z)$

Q6 (i) Write the following cubes in expanded form: $(2x + 1)^3$

Answer:

Given is $(2x + 1)^3$

We will use identity

$(a+b)^3=a^3+b^3+3a^2b+3ab^2$

Here, $a= 2x \ \ and \ \ b= 1$

Therefore,

$(2x+1)^3=(2x)^3+(1)^3+3.(2x)^2.1+3.2x.(1)^2$

$= 8x^3+1+12x^2+6x$

Q6 (ii) Write the following cube in expanded form: $(2a-3b)^3$

Answer:

Given is $(2a-3b)^3$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here, $x= 2a \ \ and \ \ y= 3b$

Therefore,

$(2a-3b)^3=(2a)^3-(3b)^3-3.(2a)^2.3b+3.2a.(3b)^2$

$= 8a^3-9b^3-36a^2b+54ab^2$

Q6 (iii) Write the following cube in expanded form: $\left[\frac{3}{2}x + 1\right ]^3$

Answer:

Given is $\left[\frac{3}{2}x + 1\right ]^3$

We will use identity

$(a+b)^3=a^3+b^3+3a^2b+3ab^2$

Here, $a= \frac{3x}{2} \ \ and \ \ b= 1$

Therefore,

$\left(\frac{3x}{2}+1 \right )^3= \left(\frac{3x}{2} \right )^3+(1)^3+3.\left(\frac{3x}{2} \right )^2.1+3.\frac{3x}{2}.(1)^2$

$= \frac{27x^3}{8}+1+\frac{27x^2}{4}+\frac{9x}{2}$

Q6 (iv) Write the following cube in expanded form: $\left[x - \frac{2}{3} y\right ]^3$

Answer:

Given is $\left[x - \frac{2}{3} y\right ]^3$

We will use identity

$(a-b)^3=a^3-b^3-3a^2b+3ab^2$

Here, $a=x \ and \ \ b= \frac{2y}{3}$

Therefore,

$\left[x - \frac{2}{3} y\right ]^3 = x^3-\left(\frac{2y}{3} \right )^3-3.x^2.\frac{2y}{3}+3.x.\left(\frac{2y}{3} \right )^2$

$= x^3-\frac{8y^3}{27}-2x^2y+\frac{4xy^2}{3}$

Q7 (i) Evaluate the following using suitable identities: $(99)^3$

Answer:

We can rewrite $(99)^3$ as

$\Rightarrow (99)^3=(100-1)^3$

We will use identity

$(a-b)^3=a^3-b^3-3a^2b+3ab^2$

Here, $a=100 \ and \ \ b= 1$

Therefore,

$(100-1)^1=(100)^3-(1)^3-3.(100)^2.1+3.100.1^2$

$= 1000000-1-30000+300= 970299$

Q7 (ii) Evaluate the following using suitable identities: $(102)^3$

Answer:

We can rewrite $(102)^3$ as

$\Rightarrow (102)^3=(100+2)^3$

We will use identity

$(a+b)^3=a^3+b^3+3a^2b+3ab^2$

Here, $a=100 \ and \ \ b= 2$

Therefore,

$(100+2)^1=(100)^3+(2)^3+3.(100)^2.2+3.100.2^2$

$= 1000000+8+60000+1200= 1061208$

Q7 (iii) Evaluate the following using suitable identities: $(998)^3$

Answer:

We can rewrite $(998)^3$ as

$\Rightarrow (998)^3=(1000-2)^3$

We will use identity

$(a-b)^3=a^3-b^3-3a^2b+3ab^2$

Here, $a=1000 \ and \ \ b= 2$

Therefore,

$(1000-2)^1=(1000)^3-(2)^3-3.(0100)^2.2+3.1000.2^2$

$= 1000000000-8-6000000+12000= 994011992$

Q8 (i) Factorise the following: $8a^3 + b^3 + 12a^2 b + 6ab^2$

Answer:

We can rewrite $8a^3 + b^3 + 12a^2 b + 6ab^2$ as

$\Rightarrow 8a^3 + b^3 + 12a^2 b + 6ab^2$ $= (2a)^3+(b)^3+3.(2a)^2.b+3.2a.(b)^2$

We will use identity

$(x+y)^3=x^3+y^3+3x^2y+3xy^2$

Here, $x=2a \ \ and \ \ y= b$

Therefore,

$8a^3 + b^3 + 12a^2 b + 6ab^2 = (2a+b)^3$

$=(2a+b)(2a+b)(2a+b)$

Q8 (ii) Factorise the following: $8a ^3 - b^3 - 12a^2 b + 6ab^2$

Answer:

We can rewrite $8a ^3 - b^3 - 12a^2 b + 6ab^2$ as

$\Rightarrow 8a ^3 - b^3 - 12a^2 b + 6ab^2$ $= (2a)^3-(b)^3-3.(2a)^2.b+3.2a.(b)^2$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here, $x=2a \ \ and \ \ y= b$

Therefore,

$8a ^3 - b^3 - 12a^2 b + 6ab^2 =(2a-b)^3$

$=(2a-b)(2a-b)(2a-b)$

Q8 (iii) Factorise the following: $27 - 125a^ 3 - 135a + 225a^2$

Answer:

We can rewrite $27 - 125a^ 3 - 135a + 225a^2$ as

$= (3)^3-(25a)^3-3.(3)^2.5a+3.3.(5a)^2$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here, $x=3 \ \ and \ \ y= 5a$

Therefore,

$27 - 125a^ 3 - 135a + 225a^2 = (3-5a)^3$

$=(3-5a)(3-5a)(3-5a)$

Q8 (iv) Factorise the following: $64a^3 - 27b^3 - 144a^2 b + 108ab^2$

Answer:

We can rewrite $64a^3 - 27b^3 - 144a^2 b + 108ab^2$ as

$\Rightarrow 64a^3 - 27b^3 - 144a^2 b + 108ab^2$ $= (4a)^3-(3b)^3-3.(4a)^2.3b+3.4a.(3b)^2$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here, $x=4a \ \ and \ \ y= 3b$

Therefore,

$64a^3 - 27b^3 - 144a^2 b + 108ab^2=(4a-3b)^2$

$=(4a-3b)(4a-3b)(4a-3b)$

Q8 (v) Factorise the following: $27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$

Answer:

We can rewrite $27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$ as

$\Rightarrow 27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$ $= (3p)^3-\left(\frac{1}{6} \right )^3-3.(3p)^2.\frac{1}{6}+3.3p.\left(\frac{1}{6} \right )^2$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here, $x=3p \ \ and \ \ y= \frac{1}{6}$

Therefore,

$27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p = \left ( 3p-\frac{1}{6} \right )^3$

$= \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right )$

Q9 (i) Verify: $x^3 + y^3 = (x +y)(x^2 - xy + y^2)$

Answer:

We know that

$(x+y)^3=x^3+y^3+3xy(x+y)$

Now,

$\Rightarrow x^3+y^3=(x+y)^3-3xy(x+y)$

$\Rightarrow x^3+y^3=(x+y)\left((x+y)^2-3xy \right )$

$\Rightarrow x^3+y^3=(x+y)\left(x^2+y^2+2xy-3xy \right )$ $(\because (a+b)^2=a^2+b^2+2ab)$

$\Rightarrow x^3+y^3=(x+y)\left(x^2+y^2-xy \right )$

Hence proved.

Q9 (ii) Verify: $x^3 - y^3 = (x -y)(x^2 + xy + y^2)$

Answer:

We know that

$(x-y)^3=x^3-y^3-3xy(x-y)$

Now,

$\Rightarrow x^3-y^3=(x-y)^3+3xy(x-y)$

$\Rightarrow x^3-y^3=(x-y)\left((x-y)^2+3xy \right )$

$\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2-2xy+3xy \right )$ $(\because (a-b)^2=a^2+b^2-2ab)$

$\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2+xy \right )$

Hence proved.

Q10 (i) Factorise the following: $27y^3 + 125z^3$

Answer:

We know that

$a^3+b^3=(a+b)(a^2+b^2-ab)$

Now, we can write $27y^3 + 125z^3$ as

$\Rightarrow 27y^3 + 125z^3 = (3y)^3+(5z)^3$

Here, $a = 3y \ \ and \ \ b = 5z$

Therefore,

$27y^3+125z^3= (3y+5z)\left((3y)^2+(5z)^2-3y.5z \right )$

$27y^3+125z^3= (3y+5z)\left(9y^2+25z^2-15yz \right )$

Q10 (ii) Factorise the following: $64m^3 - 343n^3$

Answer:

We know that

$a^3-b^3=(a-b)(a^2+b^2+ab)$

Now, we can write $64m^3 - 343n^3$ as

$\Rightarrow 64m^3 - 343n^3 = (4m)^3-(7n)^3$

Here, $a = 4m \ \ and \ \ b = 7n$

Therefore,

$64m^3-343n^3= (4m-7n)\left((4m)^2+(7n)^2+4m.7n \right )$

$64m^3-343n^3= (4m-7n)\left(16m^2+49n^2+28mn \right )$

Q11 Factorise: $27x^3 + y^3 + z^3 - 9xyz$

Answer:

Given is $27x^3 + y^3 + z^3 - 9xyz$

Now, we know that

$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Now, we can write $27x^3 + y^3 + z^3 - 9xyz$ as

$\Rightarrow 27x^3 + y^3 + z^3 - 9xyz$ $=(3x)^3+(y)^3+(z)^3-3.3x.y.z$

Here, $a= 3x , b = y \ \ and \ \ c = z$

Therefore,

$27x^3 + y^3 + z^3 - 9xyz$ $=(3x+y+z)\left((3x)^2+(y)^2+(z)^2-3x.y-y.z-z.3x \right )$

$=(3x+y+z)\left(9x^2+y^2+z^2-3xy-yz-3zx \right )$

Q12 Verify that $x^3 + y^3 + z^3 -3xyz = \frac{1}{2} ( x + y + z)\left[(x-y)^2 + (y-z)^2 + (z-x)^2 \right ]$

Answer:

We know that

$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Now, multiply and divide the R.H.S. by 2

$x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)(2x^2+2y^2+2z^2-2xy-2yz-2zx)$

$= \frac{1}{2}(x+y+z)(x^2+y^2-2xy+x^2+z^2-2zx+y^2+z^2-2yz)$

$= \frac{1}{2}(x+y+z)\left((x-y)^2+(y-z)^2 +(z-x)^2\right )$ $\left(\because a^2+b^2-2ab=(a-b)^2 \right )$

Hence proved.

Q13 If $x + y + z = 0$ , show that $x^3 + y^3 + z^3 = 3xyz$ .

Answer:

We know that

$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Now, It is given that $x + y + z = 0$

Therefore,

$x^3+y^3+z^3-3xyz =0(x^2+y^2+z^2-xy-yz-zx)$

$x^3+y^3+z^3-3xyz =0$

$x^3+y^3+z^3=3xyz$

Hence proved.

Q14 (i) Without actually calculating the cubes, find the value of each of the following: $(-12)^3 + (7)^3 + (5)^3$

Answer:

Given is $(-12)^3 + (7)^3 + (5)^3$

We know that

If $x+y+z = 0$ then , $x^3+y^3+z^3 = 3xyz$

Here, $x = -12 , y = 7 \ \ an d \ \ z = 5$

$\Rightarrow x+y+z = -12+7+5 = 0$

Therefore,

$(-12)^3 + (7)^3 + (5)^3 = 3 \times (-12)\times 7 \times 5 = -1260$

Therefore, value of $(-12)^3 + (7)^3 + (5)^3$ is $-1260$

Q14 (ii) Without actually calculating the cubes, find the value of the following: $(28)^3 + (-15)^3 + (-13)^3$

Answer:

Given is $(28)^3 + (-15)^3 + (-13)^3$

We know that

If $x+y+z = 0$ then , $x^3+y^3+z^3 = 3xyz$

Here, $x = 28 , y = -15 \ \ an d \ \ z = -13$

$\Rightarrow x+y+z =28-15-13 = 0$

Therefore,

$(28)^3 + (-15)^3 + (-13)^3 = 3 \times (28)\times (-15) \times (-13) = 16380$

Therefore, value of $(28)^3 + (-15)^3 + (-13)^3$ is $16380$

Q15 (i) Give possible expressions for the length and breadth of the following rectangle, in which its area is given:

Answer:

We know that

Area of rectangle is = $length \times breadth$

It is given that area = $25a^2-35a+12$

Now, by splitting middle term method

$\Rightarrow 25a^2-35a+12 = 25a^2-20a-15a+12$

$= 5a(5a-4)-3(5a-4)$

$= (5a-3)(5a-4)$
Therefore, two answers are possible

case (i) :- Length = $(5a-4)$ and Breadth = $(5a-3)$

case (ii) :- Length = $(5a-3)$ and Breadth = $(5a-4)$

Q15 (ii) Give possible expressions for the length and breadth of the following rectangle, in which its area is given:

Answer:

We know that

Area of rectangle is = $length \times breadth$

It is given that area = $35y^2 + 13y- 12$

Now, by splitting the middle term method

$\Rightarrow 35y^2 + 13y- 12 =35y^2+28y-15y-12$

$= 7y(5y+4)-3(5y+4)$

$= (7y-3)(5y+4)$

Therefore, two answers are possible

case (i) :- Length = $(5y+4)$ and Breadth = $(7y-3)$

case (ii) :- Length = $(7y-3)$ and Breadth = $(5y+4)$

Q16 (i) What are the possible expressions for the dimensions of the cuboid whose volumes is given below?

Answer:

We know that

Volume of cuboid is = $length \times breadth \times height$

It is given that volume = $3x^2-12x$

Now,

$\Rightarrow 3x^2-12x=3\times x\times (x-4)$

Therefore,one of the possible answer is possible

Length = $3$ and Breadth = $x$ and Height = $(x-4)$

Q16 (ii) What are the possible expressions for the dimensions of the cuboid whose volumes is given below?

Answer:

We know that

Volume of cuboid is = $length \times breadth \times height$

It is given that volume = $12ky^2+8ky-20k$

Now,

$\Rightarrow 12ky^2+8ky-20k = k(12y^2+8y-20)$

$= k(12y^2+20y-12y-20)$

$= k\left(4y(3y+5)-4(3y+5) \right )$

$= k(3y+5)(4y-4)$

$= 4k(3y+5)(y-1)$

Therefore,one of the possible answer is possible

Length = $4k$ and Breadth = $(3y+5)$ and Height = $(y-1)$