Here a3 – 8 can be written as a3 – 23
Comparing this with a3−b3, we get a=a,b=2
\(\therefore\) a³-b³ = (a-b)(a²+ab+b²)
a³-2³=(a-2)(a²+a(2)+22)
a³-8=(a-2)(a²+2a+4)

Comparing x³+ 5³ with a³+b³ we get a = x, b = 5
We know,
a³+ b³ = (a+b) (a²−ab+b²)
x³+ 5³ = (x+5) (x²−(x)(5)+5²)
x³+ 5³ = (x+5) (x²−5x+25)

This is in the form of ax² + bx + c
We get a = 1, b = 8, c = 15
Now, the product = a x c and sum = b
= 1 x 15 b = 8
= x²+ 8x +15
= (x² + 3x) + (5x + 15) (the middle term 8x can be written as 3x + 5x)
= x(x + 3) + 5(x + 3) (taking out the common factor x + 3 )
x² + 8x + 15 = (x + 5) (x + 3)

Now, 49x² - 84xy + 36y² = 72x² - 84xy + 6²y²
= (7x)² - 2(7x)(6y) + (6y)²
Comparing this with a² - 2ab + b² = (a - b)²
(7x)²- 2(7x)(6y) + (6y)² = (7x - 6y)²
\(\therefore\) 49x²- 84xy + 36y² = (7x-6y)²

We have, x² + yz + xy + xz
Group the terms suitably as = (x² + xy) + (yz + xz)
= x(x + y) + z(y + x) (addition is commutative)
= x(x + y) + z(x + y) [taking out the common factor(x + y)]

We have,4x² y + 8xy
This can be written as = (2 x 2 X x X x y )+ (2 X 2 X x X y)
Taking out the common factor 2,2,x,y,we get
= 2 x 2 X x X y(x + 2)
= 4xy (x + 2)
= 4xy (x + 2)

Given:
MA = 4 cm, AT = 3.6 cm,
TH = 4.5 cm, MH = 5 cm and ∠A = 85°


Steps:
1. Draw a line segment MA = 4 cm.
2. Make ∠A = 85°.
3. With A as centre, draw an arc of radius 3.6 cm. Let it cut the ray AX at T.
4. With M and T as centres, draw arcs of radii 5 cm and 4.5 cm respectively and let them cut at H.
5. Join MH and TH.
6. MATH is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral MATH = \(\frac12\) × d × (h₁+ h₂) sq.units
= \(\frac12\) x 5.1 x (3.9 + 2.8)
= 2.55 x 6.7 = 17.09 cm²

Given: DE = 6 cm, EA = 5 cm, AR = 5.5 cm,
RD = 5.2 cm and a diagonal DA = 10 cm


Steps:
1. Draw a line segment DE = 6 cm.
2. With D and E as centres, draw arcs of radii 10 cm and 5 cm respectively and let them cut at A.
3. Join DA and EA.
4. With D and A as centres, draw arcs of radii 5.2 cm and 5.5 cm respectively and let them cut at R.
5. Join DR and AR.
6. DEAR is the required quadrilateral
Calculation of Area:
Area of the quadrilateral DEAR = \(\frac12\) x d x (h₁+ h₂) sq. units
= \(\frac12\) x 10 x (1.9+ 2.3)
= 5 x 4.2 = 21 cm²

We have = \(\cfrac { { 10m }^{ 2 }-5m }{ (2m-1) } \)
 (
(taking common factor from the numerator)
= 5m

We have, (5y³ − 25y² + 8y) ÷ 5y
= \(\cfrac { { 5y }^{ 3 }-25{ y }^{ 2 }+8y }{ 5y } \)

 = \({ y }^{ 3-1 }-{ 5y }^{ 2-1 }+\cfrac { 8 }{ 5 } \)
= \({ y }^{ 2 }-5y+\cfrac { 8 }{ 5 } \)

The given field has four triangles (I, III, IV & V) and a trapezium (II).
Area of the triangle (I) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 5\times 6=15{ m }^{ 2 }\)
Area of the trapezium (II) \(=\frac { 1 }{ 2 } h(a+b)=\frac { 1 }{ 2 } \times 13\times (6+4)=65{ m }^{ 2 }\)
Area of the triangle (III) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 8\times 4=\frac { 32 }{ 2 } =16{ m }^{ 2 }\)
Area of the triangle (IV) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 13\times 10=65{ m }^{ 2 }\)
Area of the triangle (V) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 13\times 10=65{ m }^{ 2 }\)
∴ The total area of the field = 15 + 65 + 16 + 65 + 65 = 226 m²

The mat given in the figure can be split into two rectangles as follows
∴ Area of the entire mat
= area of the I rectangle + area of the II rectangle
= (l₁ x b₁) + (l₂ x b₂)
= 5 x 2 + 9 x 2 = 10 + 18 = 28 sq.feet
Cost per sq. foot = Rs. 20
∴ The total cost of the entire mat = 28 x Rs. 20 = Rs. 560.

The door mat has two semi-circles attached to a rectangle.
Length of the rectangle = 90 – 40 = 50 cm
Its breadth = 40 cm
Diameter of the semi-circle = 40 cm
Its radius = 20 cm
∴ Area of the door mat = area of the rectangle + 2 × area of the semicircle
\(=(l\times b)+\left( 2+\frac { 1 }{ 2 } { \pi r }^{ 2 } \right) \)
= (50 x 40) + (3.14 x 20 x 20)
= 2000 + 1256 = 3256 sq.cm (approximately)

The door is made of semicircle and rectangular shapes.
Length of the rectangle = 9 –3.5 = 5.5 feet
Its breadth = 3.5 feet
Diameter of the semicircle = 3.5 feet
∴ Radius =\(\frac{3.5}{2}=1.75\)feet
∴ Area of the door = Area of the rectangle + Area of the semicircle
\(=(l\times b)+\left( \frac { 1 }{ 2 } { \pi r }^{ 2 } \right) \)
\(=(5.5\times 3.5)+\left( \frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times 1.75\times 1.75 \right) \)
= 19.25 + 4.81 = 24.06 sq.feet (approx.)

Radius of a circular quadrant, r = 3.5 cm and side of a square, a = 3.5 cm.
The given figure is formed by the joining of 4 quadrants of a circle with each side of a square. The boundary of the given figure consists of 4 arcs and 4 radii.
(i) Perimeter of the given combined shape
= 4 x length of the arcs of the quadrant of a circle + 4 x radius
\(=\left( 4\times \frac { 1 }{ 4 } \times 2\pi r \right) +4r\)
\(=\left( 4\times \frac { 1 }{ 4 } \times 2\times 3.5 \right) +(1\times 3.5)\)
= 22 + 14 = 36 cm (approximately)
(ii) Area of the given combined shape
= area of the square + 4 x area of the quadrants of the circle
\({ a }^{ 2 }=\left( 4\times \frac { 1 }{ 4 } \times \pi { r }^{ 2 } \right) \)
\(=(3.5\times 3.5)+\left( \frac { 22 }{ 7 } \times 3.5\times 3.5 \right) \)
A = 12.25 + 38.5 = 50.75 cm² (approximately)